\[TOC\] # 数论模板 ## 一, 线性筛 所有数组定义在外面, 自动置为全0即可 
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void init() {
    cnt = 0; //统计素数个数
    is_prime[1] = 1; //0代表是素数, 1不是素数
    for (int i = 2; i <= maxn; i++) {
        if (!is_prime[i]) {
            prime[++cnt] = i; //i是素数, prime从下标[1]开始有记录
        }
        for (int j = 1; j <= cnt; j++) { // 1 到 cnt !
            if (prime[j] * i >= maxn)
                break;
            is_prime[i * prime[j]] = 1;
            if (i % prime[j] == 0)
                break;
        }
    }
}
### 1.分解质因数 通过素数筛法打表, 然后只能通过优化暴力, **最快了** 
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int get_primefactors(int n) {
    int sum = 0;
    for (int i = 1; i <= tot && pri[i] * pri[i] <= n; ++i) {
        if (n % pri[i] == 0) {
            fac[++sum] = pri[i];
            while (n % pri[i] == 0) {
                n /= pri[i];
            }
        }
    }
    if (n > 1) {
        fac[++sum] = n;
    }
    return sum;
}
### 2.区间线性筛 a, b可以很大, 但是b - a 的结果可以接受 模板 : 输出a 到 b之间的素数个数 
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#define ll long long
const int MAXN = 100005;
bool is[MAXN];
int pri[MAXN / 10];
int cnt;
bool ans[MAXN];

void init() {
    memset(is, true, sizeof(is));
    is[0] = is[1] = false;
    cnt = 0;
    for (int i = 2; i < MAXN; i++) {
        if (is[i]) {
            pri[++cnt] = i;
        }
        for (int j = 1; j < cnt && pri[j] * i < MAXN; j++) {
            is[pri[j] * i] = false;
            if (i % pri[j] == 0) {
                break;
            }
        }
    }
}

int solve(ll a, ll b) {
    memset(ans, true, sizeof(ans));
    for (int i = 1; i <= cnt && 1ll * pri[i] * pri[i] <= b; i++) {
        ll s = (a + pri[i] - 1) / pri[i];
        if (s < 2) {
            s = 2;
        }
        s *= pri[i];
        for (; s <= b; s += pri[i]) {
            ans[s - a] = false;
        }
    }
    int res = 0;
    for (int i = 0; i <= b - a; i++) {
        if (ans[i]) {
            res++;
        }
    }
    if (a == 1) {
        res--;
    }
    return res;
}

int main() {
    init();
    int a, b;
    cin >> a >> b;
    cout << solve(a, b);
    // 2 , 3 , 5,  7
}
### 3.求各种积性函数 YC究极版! 
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#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 1e7 + 5;
bool is[N]; //筛数组
int pri[N]; //素数数组
int phi[N]; //欧拉函数数组 !
int mu[N]; //莫⽐乌斯函数数组
int e[N]; //n的最⼩素因⼦在唯⼀分解中的幂次
int d[N]; //因数个数 !
int fac[N]; //i的最⼩的幂次⼤于1的质因数的幂
int tot; //素数个数

void init() {
    memset(is, true, sizeof(is));
    tot = 0;
    is[0] = is[1] = 0;
    mu[1] = 1;
    phi[1] = 1;
    fac[1] = 1;
    e[1] = 0;
    d[1] = 1;
    for (int i = 2; i < N; i++) {
        if (is[i]) {
            pri[++tot] = i;
            phi[i] = i - 1;
            mu[i] = -1;
            e[i] = 1;
            d[i] = 2;
            fac[i] = i;
        }
        for (int j = 1; j <= tot && pri[j] * i < N; j++) {
            is[pri[j] * i] = 0;
            if (i % pri[j] == 0) {
                phi[i * pri[j]] = phi[i] * pri[j];
                mu[i * pri[j]] = 0;
                d[pri[j] * i] = d[i] / (e[i] + 1) * (e[i] + 2);
                e[pri[j] * i] = e[i] + 1;
                fac[i * pri[j]] = fac[i] * pri[j];
                break;
            }
            else {
                mu[i * pri[j]] = -mu[i];
                phi[i * pri[j]] = phi[i] * (pri[j] - 1);
                e[pri[j] * i] = 1;
                d[pri[j] * i] = d[i] * d[pri[j]];
                fac[i * pri[j]] = pri[j];
            }
        }
    }
}


int main() {
    init();
    int a, b;
    cin >> a >> b;
    cout << d[a] << ' ' << d[b];
    return 0;
}
## 二, Min\_25筛 #### 求十亿(1e9)内所有质数的和 $O(n ^{3/4})$ 
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// 求比n小的所有素数的和
const int N = 1000010;
typedef long long LL;

namespace Min25 {

    int prime[N], id1[N], id2[N], flag[N], ncnt, m;

    LL g[N], sum[N], a[N], T, n;

    inline int ID(LL x) {
        return x <= T ? id1[x] : id2[n / x];
    }

    inline LL calc(LL x) {
        return x * (x + 1) / 2 - 1;
    }

    inline LL f(LL x) {
        return x;
    }

    inline void init() {
        T = sqrt(n + 0.5);
        for (int i = 2; i <= T; i++) {
            if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
            for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) {
                flag[i * prime[j]] = 1;
                if (i % prime[j] == 0) break;
            }
        }
        for (LL l = 1; l <= n; l = n / (n / l) + 1) {
            a[++m] = n / l;
            if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
            g[m] = calc(a[m]);
        }
        for (int i = 1; i <= ncnt; i++)
            for (int j = 1; j <= m && (LL)prime[i] * prime[i] <= a[j]; j++)
                g[j] = g[j] - (LL)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
    }

    inline LL solve(LL x) {
        if (x <= 1) return x;
        return n = x, init(), g[ID(n)];
    }

}

int main() {
    LL n; scanf("%lld", &n);
    printf("%lld\n", Min25::solve(n));
}
#### 求积性函数的前缀和 
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#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 2000000+100;

/******************************
f()函数中(31-37行) 填函数在质数幂次处的表达式
pow_sum()函数中(38-43行) 填幂和函数（如果需要更高次的话可以在这里添加）
202-205行按要求填写
f_p[][0/1/2/3/...]分别代表质数个数/质数和/质数平方和/质数三次方和/...根据自己需要添加
例：如果该函数在质数处表达式为f(p) = p^2+3*p+1
则表明需要质数个数/质数和/质数平方和
即f_p[][0],f_p[][1],f_p[][2]
******************************/

ll poww(ll a,ll b){
    ll res = 1;
    ll base = a;
    while(b){
        if(b&1){
            res *= base;
            //res %= mod;
        }
        base *= base;
        //base %= mod;
        b>>=1;
    }
    return res;
}

inline ll f(ll p,int e){
    if(p==1||e==0) return 1;
    ///return f(p^e)
    ll res = poww(p,e);
    return res*res+3*res+1;

}
ll pow_sum(ll n,int k){
    ///return sum(i^k),i from 1 to n.
    if(k==0) return n;
    if(k==1) return n*(n+1)/2;
    if(k==2) return n*(n+1)*(2*n+1)/6;
}
ll f_p[maxn][3];///F_prime(id(n/i))
ll n;
int n_2; ///(int)sqrt(n)
int n_3; ///(int)pow(n,1.0/3.0)
int n_6; ///(int)pow(n,1.0/6.0)
ll val_id[maxn]; ///give the id, return the id-th number like 'n/i' ,(val_id[1] = 1)
int val_id_num; ///how many numbers like 'n/i'
int val_id_num_3; ///how many numbers like 'n/i' below n/n_3;
int p[200000+100];
bool isp[maxn];
int p_sz_2; ///pi(n_2)
int p_sz_3; ///pi(n_3)
int p_sz_6; ///pi(n_6)
void init(){
    n_2 = (int)sqrt(n);
    n_3 = (int)pow(n,1.0/3.0);
    n_6 = (int)pow(n,1.0/6.0);
    val_id_num = 0;
    for(ll i=1;i<=n;){
        val_id[++val_id_num] = i;
        if(i==n) break;
        i = n/(n/(i+1));
    }
    memset(isp,1,sizeof isp);
    isp[1] = 0;
    for(int i=2;i<=n_2;i++){
        if(isp[i]){
            p[++p_sz_2] = i;
            if(i<=n_3) p_sz_3++;
            if(i<=n_6) p_sz_6++;
        }
        for(int j=1;j<=p_sz_2&&p[j]*i<=n_2;j++){
            isp[i*p[j]] = 0;
            if(i%p[j]==0) break;
        }
    }
}
inline int get_id(ll k){ ///give a number like 'n/i', return the id of it
    if(k>n_2) return val_id_num-n/k+1;
    else return k;
}
ll c[maxn];
int lowbit(int n){return n & (-n);}
void add(int x,ll d){
    while(x<maxn){
        c[x]+=d;
        x+=lowbit(x);
    }
}
ll sum(int x){
    ll ans=0;
    while(x){
        ans+=c[x];
        x-=lowbit(x);
    }
    return ans;
}

struct node{
    int k_max;
    ll val;
    ll f_val;
};
void update_bfs(int k,int type){
    queue<node> q;
    while(!q.empty()) q.pop();
    int e = 1;
    for(ll i=p[k];i<n/n_3;i*=p[k]){
        node st;
        st.k_max = k;
        st.val = i;
        if(type==-1)st.f_val = f(p[k],e);
        else st.f_val = poww(i,type);
        q.push(st);
        e++;
    }
    while(!q.empty()){
        node hd = q.front();
        q.pop();
        if((hd.val!=p[hd.k_max]&&type>=0)||type==-1) {//if(type==-1)cout << "****" << hd.val << "****" << hd.f_val << endl;
            ll w = n/hd.val;
            w = n/w;//cout << hd.val << "[" << w<<" , " << val_id[val_id_num] << "]" << endl;
            if(type==-1){
                add(get_id(w),hd.f_val);
                add(val_id_num+1,-1ll*hd.f_val);
            }
            else{
                add(get_id(w),-1ll*hd.f_val);
                add(val_id_num+1,hd.f_val);
            }
        }
        for(int i=hd.k_max+1;hd.val*p[i]<n/n_3&&i<=p_sz_2;i++){
            ll res = p[i];
            for(int e=1;;e++){
                if(hd.val*res<n/n_3){
                    node nxt;
                    nxt.k_max = i;
                    nxt.val = hd.val*res;
                    if(type==-1) nxt.f_val = hd.f_val*f(p[i],e);
                    else nxt.f_val = hd.f_val*poww(res,type);
                    q.push(nxt);
                }
                else break;
                res *= p[i];
            }
        }
    }
}
void get_f_p(ll n,int times){
    for(int i=1;i<=val_id_num;i++){
        for(int j=0;j<=times;j++){
            f_p[i][j] = pow_sum(val_id[i],j)-1;
        }
    }
    int now;
    //for(now=1;now<=p_sz_2;now++){
    for(now=1;p[now]<=n_6;now++){
        for(int j=val_id_num;j>=1;j--){
            ll w = val_id[j]/p[now];
            if(w<p[now]) break;
            ll val=1;
            for(int k = 0;k<=times;k++){
                f_p[j][k] = f_p[j][k] - val*(f_p[get_id(w)][k]-f_p[p[now-1]][k]);
                val *= p[now];
            }
        }
    }
    int nnow = now;
    int val = 1;
    for(int tt = 0;tt<=times;tt++){
        now = nnow;
        memset(c,0,sizeof c);
        add(1,f_p[1][tt]);
        for(int i=2;val_id[i]<n/n_3;i++){
            add(i,f_p[i][tt] - f_p[i-1][tt]);
        }
        for(;p[now]<=n_3;now++){
            for(int j=val_id_num;j>=1;j--){
                ll w = val_id[j]/p[now];
                if(val_id[j]<n/n_3) break;
                if(w<p[now]) break;
                if(w<n/n_3) f_p[j][tt] = f_p[j][tt] - (sum(get_id(w)) - sum(p[now-1]))*poww(p[now],tt);
                else f_p[j][tt] = f_p[j][tt] - (f_p[get_id(w)][tt]-sum(p[now-1]))*poww(p[now],tt);
            }
            update_bfs(now,tt);
        }
        for(int i=1;i<=val_id_num&&val_id[i]<n/n_3;i++)
            f_p[i][tt] = sum(i);
        for(;now<=p_sz_2;now++){
            for(int j=val_id_num;j>=1;j--){
                ll w = val_id[j]/p[now];
                if(val_id[j]<n/n_3) break;
                if(w<p[now]) break;
                f_p[j][tt] -= (f_p[get_id(w)][tt]-f_p[p[now-1]][tt])*poww(p[now],tt);
            }
        }
    }

    for(int i=1;i<=val_id_num;i++){
        ///if f(p) = p^2+3p+1,then write:f_p[i][0] = f_p[i][2] + 3*f_p[i][1] + f_p[i][0];
        f_p[i][0] = f_p[i][2] + 3*f_p[i][1] + f_p[i][0];
    }

}
ll F[2000000+100];
void get_f_3(ll n){ ///V(F_{pi(n^(1/3))+1},n)
    ll q = p[p_sz_3+1];
    for(int now=1;now<=val_id_num;now++){
        if(val_id[now]<q){
            F[now] = 1;
        }
        else if(val_id[now]<q*q){
            F[now] = 1+(f_p[now][0]-f_p[q-1][0]);
        }
        else{
            F[now] = 1+(f_p[now][0]-f_p[q-1][0]);
            for(int pp=p_sz_3+1;p[pp]<=(int)(sqrt(val_id[now]))&&pp<=p_sz_2;pp++){
                F[now] += f(p[pp],2) + (f(p[pp],1))*(f_p[get_id(val_id[now]/p[pp])][0]-f_p[get_id(p[pp])][0]);
            }
        }
    }
}
void get_f_6(ll n){ ///V(F_{pi(n^(1/6))+1},n)
    memset(c,0,sizeof c);
    add(1,F[1]);
    for(int i=2;val_id[i]<n/n_3;i++){
        add(i,F[i] - F[i-1]);
    }
    for(int k=p_sz_3;k>p_sz_6;k--){
        int now = val_id_num;
        for(;val_id[now]>=n/n_3;now--){
            int e = 1;
            ll _p = p[k];
            while(val_id[now]/_p){
                if(val_id[now]/_p>=n/n_3){
                    F[now] += F[get_id(val_id[now]/_p)]*f(p[k],e);
                }
                else{
                    F[now] += sum(get_id(val_id[now]/_p))*f(p[k],e);
                }
                _p *= p[k];
                e++;
            }
        }
        if(k==1) break;
        //cout << "******" << p[k] << "******" << n/n_3 << endl;
        update_bfs(k,-1);///bfs to update [lpf(i)==P{k-1}]f(i)
    }
    for(int i=1;i<=val_id_num&&val_id[i]<n/n_3;i++)
        F[i] = sum(i);
}
void get_f(ll n){
    for(int k=p_sz_6;k>=1;k--){
        for(int now = val_id_num;now>=1;now--){
            int e = 1;
            ll _p = p[k];
            while(val_id[now]/_p){
                    F[now] += F[get_id(val_id[now]/_p)]*f(p[k],e);
                _p *= p[k];
                e++;
            }
        }
    }
}
int main(){//n = 1000000000; //1e10:455052511,0.83s/0.58s 1e12:37607912018 9.224s/5.105s
    cin >> n;
    init();
    get_f_p(n,2);
    get_f_3(n);
    get_f_6(n);
    get_f(n);
    for(int i=1;i<=val_id_num;i++){
        cout << val_id[i] << " : " << F[i] << endl;
    }

}

```  

## 三, 唯一分解定理(算术基本定理)

暴力分解即可

代码没有

## 四, GCD/EXGCD

用途: **ax + by = gcd(a,b) 方程求解** 

```C++
#include <bits/stdc++.h>
using namespace std;
int a, b, x, y；

void EXGCD(int a, int b, int &x, int &y) {
    if (!b) { x = 1; y = 0; return; }
    EXGCD(b, a%b, y, x);
    y -= a / b * x;
}

int main() {
    scanf("%d %d", &a, &b);
    EXGCD(a,b,x,y);
    printf("%d %d", x, y);
    return 0;
}
使用STL的代码: 
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#include <bits/stdc++.h>
using namespace std;
#define pir pair<int,int>    
int a, b, x, y;

pir EXGCD(int a,int b) {
    if (!b) return make_pair(1, 0);
    pir tmp = EXGCD(b, a%b);
    return make_pair(tmp.second ,tmp.first - a/b*tmp.second);
}

int main() {
    scanf("%d %d", &a, &b);
    pir ans = EXGCD(a,b);
    printf("%d %d",ans.first,ans.second);
    return 0;
}
## 五, 中国剩余定理(孙子定理) ### 1.CRT(保证$m\_1,m\_2…m\_n$之间两两互质，求最小的_x_) 用途: **同余方程组求解!** ![image-20201006204034516](C:%5CUsers%5CDELL%5CAppData%5CRoaming%5CTypora%5Ctypora-user-images%5Cimage-20201006204034516.png) 我们求出了任意一个满足条件的_x_之后，只需要对其模_M_就是最终的答案(因为_M_M是所有数的_lcm_) 
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#include<cstdio>
#define ll long long
//扩展欧几里得算法 
void gcd(ll a,ll b,ll &d,ll &x,ll &y)
{
    if(b==0){
        d=a;
        x=1,y=0;
    }
    else{//else不能省略 
        gcd(b,a%b,d,y,x);
        y-=(a/b)*x;
    }
}
//中国剩余定理 
ll China(int n,ll *m,ll *a)
{
    ll M=1,d,y,x=0;
    for(int i=0;i<n;i++) M*=m[i];
    for(int i=0;i<n;i++){
        ll w=M/m[i];
        gcd(m[i],w,d,d,y);
        x=(x+y*w*a[i])%M;
    }
    return (x+M)%M;
}
### 2\. EXCRT($m\_1,m\_2…m\_n$之间不再限制互质，求最小的_x_) 
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#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=100000+10;
int n;
ll a[N],r[N];
ll exgcd(ll a,ll b,ll &x,ll &y){
    if(b==0){
        x=1,y=0;return a;
    }    
    ll ret=exgcd(b,a%b,y,x);
    y-=(a/b)*x;
    return ret;
}
ll excrt(){
    ll M=a[1],R=r[1],x,y,d;
    for(int i=2;i<=n;i++){
        d=exgcd(M,a[i],x,y);
        if((R-r[i])%d) return -1;
        x=(R-r[i])/d*x%a[i];
        R-=M*x;
        M=M/d*a[i];
        R%=M;
    }
    return (R%M+M)%M;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%lld%lld",&a[i],&r[i]);
        }
        printf("%lld\n",excrt());
    }
    return 0;
}
## 六, 米勒罗宾\_大素数测试 
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#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int S = 8;

//计算ret = (a*b)%c a, b, c < 2^63
ll mult_mod(ll a, ll b, ll c) {
    a %= c;
    b %= c;
    ll ans = 0;
    ll tmp = a;
    while (b) {
        if (b & 1) {
            ans += tmp;
            if (ans > c) {
                ans -= c;
            }
        }
        tmp <<= 1;
        if (tmp > c) {
            tmp -= c;
        }
        b >>= 1;
    }
    return ans;
}
//计算 ret = (a^n) % mod
ll pow_mod(ll a, ll n, ll mod) {
    ll ans = 1;
    ll tmp = a % mod;
    while (n) {
        if (n & 1) {
            ans = mult_mod(ans, tmp, mod);
        }
        tmp = mult_mod(tmp, tmp, mod);
        n >>= 1;
    }
    return ans;
}

//通过 a^(n-1) = 1(mod n) 判断n是不是素数
//n - 1 = x * 2 ^ t 中间使用二次判断
//是合数返回ture, 不一定是合数返回false

bool check(ll a, ll n, ll x, ll t) {
    ll ans = pow_mod(a, x, n);
    ll last = ans;
    for (int i = 1; i <= t; ++i) {
        ans = mult_mod(ans, ans, n);
        if (ans == 1 && last != 1 && last != n - 1) {
            return true;
        }
        last = ans;
    }
    if (ans != 1) {
        return true;
    }
    return false;
}

bool Miller_Rabin(ll n) { //是素数返回 ture, 否则返回false
    if (n < 2) {
        return false;
    }
    if (n == 2) {
        return true;
    }
    if ((n & 1) == 0) {
        return false;
    }
    ll x = n - 1;
    ll t = 0;
    t = (ll)log2(x & (-x));
    x = x / (x & (-x));
    srand(time(NULL));
    for (int i = 0; i < S; ++i) {
        ll a = rand() % (n - 1) + 1;//产生随机数a, 并控制其在0 ~ n-1之间
        if (check(a, n, x, t)) {
            return false;
        }
    }
    return true;
}
int main() {
    int t, ans;
    ll tem;
    scanf("%d", &t);
    ans = 0;
    while (t--) {
        scanf("%lld", &tem);
        if (Miller_Rabin(tem)) {
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}
### 增: Pallord\_rho 大整数分解质因数 
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#include <bits/stdc++.h>
#define ll long long

#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a, 0, sizeof(a))
const double eps = 1e-5;
using namespace std;

const int S = 30;//随机算法判定次数，S越大，判错概率越小
//计算 (a*b)%c.   a,b都是long long的数，直接相乘可能溢出的
//  a,b,c <2^63
ll mult_mod(ll a, ll b, ll c)
{
    a %= c;
    b %= c;
    ll ret = 0;
    while (b)
    {
        if (b & 1) { ret += a; ret %= c; }
        a <<= 1;
        if (a >= c)a %= c;
        b >>= 1;
    }
    return ret;
}
ll pow_mod(ll x, ll n, ll mod)
{
    if (n == 1)return x % mod;
    x %= mod;
    ll tmp = x;
    ll ret = 1;
    while (n)
    {
        if (n & 1) ret = mult_mod(ret, tmp, mod);
        tmp = mult_mod(tmp, tmp, mod);
        n >>= 1;
    }
    return ret;
}
//以a为基,n-1=x*2^t      a^(n-1)=1(mod n)  验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(ll a, ll n, ll x, ll t)
{
    ll ret = pow_mod(a, x, n);
    ll last = ret;
    for (int i = 1; i <= t; i++)
    {
        ret = mult_mod(ret, ret, n);
        if (ret == 1 && last != 1 && last != n - 1) return true;//合数
        last = ret;
    }
    if (ret != 1) return true;
    return false;
}
// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数，但概率极小)
//合数返回false;
bool Miller_Rabin(ll n)
{
    if (n < 2)return false;
    if (n == 2)return true;
    if ((n & 1) == 0) return false;//偶数
    ll x = n - 1;
    ll t = 0;
    while ((x & 1) == 0) { x >>= 1; t++; }
    for (int i = 0; i < S; i++)
    {
        ll a = rand() % (n - 1) + 1;//rand()需要stdlib.h头文件
        if (check(a, n, x, t))
            return false;//合数
    }
    return true;
}
ll factor[100];//质因数分解结果（刚返回时是无序的）
int tol;//质因数的个数。数组小标从0开始
ll gcd(ll a, ll b)
{
    if (a == 0)return 1;//???????
    if (a < 0) return gcd(-a, b);
    while (b)
    {
        ll t = a % b;
        a = b;
        b = t;
    }
    return a;
}
ll Pollard_rho(ll x, ll c)
{
    ll i = 1, k = 2;
    ll x0 = rand() % x;
    ll y = x0;
    while (1)
    {
        i++;
        x0 = (mult_mod(x0, x0, x) + c) % x;
        ll d = gcd(y - x0, x);
        if (d != 1 && d != x) return d;
        if (y == x0) return x;
        if (i == k) { y = x0; k += k; }
    }
}
//对n进行素因子分解
void findfac(ll n)
{
    if (Miller_Rabin(n))//素数
    {
        factor[tol++] = n;
        return;
    }
    ll p = n;
    while (p >= n)p = Pollard_rho(p, rand() % (n - 1) + 1);
    findfac(p);
    findfac(n / p);
}
int main() {
    ll n;
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        if (n == 1) {
            //cout << "no" << '\n';
            
            continue;
        }
        tol = 0;
        findfac(n);
        sort(factor, factor + tol); //素因子排序
        
        for (int i = 0; i < tol; i++) { //输出所有素因子!!! 有重复的哦: 100 -> 2 2 5 5
            cout << factor[i] << ' ';
        }
        //质因子
    }
    return 0;
}
## 七, 乘法逆元 ### 1.by EXGCD O(logN) 
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 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 ll exgcd(ll a,ll b,ll &x,ll &y) {
 4     if(!b) {
 5         x=1,y=0;
 6         return a;
 7     }
 8     ll res=exgcd(b,a%b,y,x);
 9     y-=a/b*x;       ///x=x1,y=x1-a/b*y1   x1,y1代表下一状态
10     return res;
11 }
12 int main()
13 {
14     ll a,p,x,y;  ///扩展欧几里得计算a的逆元（mod p）
15     scanf("%lld%lld",&a,&p);
16     ll d=exgcd(a,p,x,y);
17     printf(d==1?"%lld":"-1",(x+p)%p);///最大公约数不为1，逆元不存在，输出-1
18     return 0;
19 }
### 2.费马小定理 O(logP) 模为素数p的情况下 
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 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 ll quickpowmod(ll a,ll b,ll mod) {
 4     ll ans=1;
 5     while(b) {
 6         if(b&1) ans=(ans*a)%mod;
 7         b>>=1;
 8         a=(a*a)%mod;
 9     }
10     return ans;
11 }
12 int main()
13 {
14     ll a,p;  ///费马小定理计算a的逆元（mod p）
15     scanf("%lld%lld",&a,&p);
16     ll inva=quickpowmod(a,p-2,p);
17     printf("%lld",inva);
18     return 0;
19 }
### 3.逆元打表 
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inv[1]=1;
for(int i = 2 ; i <= n; i++)
    inv[i]=(long long)(p - p/i) * inv[p % i] % p;
## 八, 矩阵快速幂 模板是求解 : F（n）= F（n-1）+F（n-2）+F（n-3） 具体题目具体分析, 推算矩阵 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;

struct mat {
    ll a[3][3];
};

mat mat_mul(mat a,mat b) {
    mat res;
    memset(res.a,0,sizeof(res.a));
    for(int i = 0; i<=2 ; i++) {
        for(int j = 0; j<=2 ; j++) {
            for(int k = 0; k<=2 ; k++) {
                res.a[i][j] += ( a.a[i][k] * b.a[k][j])%MOD;
            }
        }
    }
    return res;
}

ll mat_pow(ll n) {
    mat c,res;
    memset(c.a,0,sizeof(c.a));
    memset(res.a,0,sizeof(res.a));
    c.a[0][0] = 1;
    c.a[0][1] = 1;
    c.a[0][2] = 1;
    c.a[1][0] = 1;
    c.a[2][1] = 1;
    for(int i=0; i<3 ; i++) {
        res.a[i][i] = 1;
    }
    while(n) {
        if(n&1) {
            res = mat_mul(res,c);
        }
        c = mat_mul(c,c);
        n >>= 1;
    }
    return (res.a[0][0])%MOD;
}

int main() {
    ll n;
    while(scanf("%lld",&n)!=EOF) {

//    n -= 1;

        ll ans = mat_pow(n);

        printf("%lld\n",ans);

    }
}
## 九, 卢卡斯定理 ### 1.Lucas (模数p是质数) 给定$n,m,p (1 <= n,m,p <= 10^5)$ , 且p为素数 求 $C^n\_{n+m} mod P$ 
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#include<iostream>
using namespace std;
typedef long long LL;
const LL N=1e5+2; //1e5 !!!

LL a[N];
void init(LL p)
{
	a[1]=1;
	for(int i=2;i<=p;++i)a[i]=a[i-1]*i%p;
}
void exgcd(LL a,LL b,LL &x,LL &y)
{
	if(!b){
		x=1;
		y=0;
		return;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
}
LL ksm(LL x,LL n,LL mod)
{
	LL ans=1;
	while(n){
		if(n&1)ans=ans*x%mod;
		n>>=1;
		x=x*x%mod;
	}
	return ans;
}
LL C(LL n,LL m,LL p)
{
	if(n==m||m==0)return 1;
	if(n<m)return 0;
	if(m*2>n)m=n-m;						  /*C(n,m)=c(n,n-m)*/
	return a[n]*ksm(a[m]*a[n-m],p-2,p)%p;
    /*求(a[m]*a[n-m])在(mod p)意义下的乘法逆元*/										/*拓展欧几里得与费马小定理均可*/ 
	/*LL x,y;
	exgcd(a[m]*a[n-m],p,x,y);
	return (a[n]*x%p+p)%p;*/ 
}
LL lucas(LL n,LL m,LL p)
{
	if(!m)return 1;
	return lucas(n/p,m/p,p)*C(n%p,m%p,p)%p;
}
int main()
{
	ios::sync_with_stdio(false);
	LL T,n,m,p;
	cin>>T;
	while(T--){
		cin>>n>>m>>p;
		init(p);
		cout<<lucas(n+m,m,p)<<endl;
	}
	return 0;
}
### 2\. EX-Lucas (模数p不一定是质数的时候) 求$C^m\_n mod P$ 
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL N=1e5+9;
LL A[N],M[N];
LL ksm(LL x,LL n,LL mod)
{
	LL ans=1;
	while(n){
		if(n&1)ans=ans*x%mod;
		n>>=1,x=x*x%mod;
	}
	return ans;
}
void exgcd(LL a,LL b,LL &x,LL &y)
{
	if(!b)x=1,y=0;
	else exgcd(b,a%b,y,x),y-=a/b*x;
}
LL inv(LL a,LL p)
{
	LL x,y;
	exgcd(a,p,x,y);
	return (x+p)%p?x:x+p;
}
LL get(LL n,LL pi,LL p)									/*求(与pi互素后的n!)%M[i]*/ 
{
	if(!n)return 1;
	LL ans=1;
	if(n/p){											/*判断有无循环节 */ 
		for(LL i=2;i<=p;++i)if(i%pi)ans=ans*i%p;
		ans=ksm(ans,n/p,p);
	}
	for(LL i=2;i<=n%p;++i)if(i%pi)ans=ans*i%p;			/*循环节剩余部分*/ 
	return ans*get(n/pi,pi,p)%p;
}
LL exlucas(LL n,LL m,LL pi,LL p)						/*求A[i]*/ 
{
	LL nn=get(n,pi,p);									/*求(与pi互素后的n)%M[i]*/ 
	LL mm=get(m,pi,p);									/*求(m!与pi互素后的m!)%M[i]*/ 
	LL nm=get(n-m,pi,p);								/*求(与pi互素后的(n-m)!)%M[i]*/ 
	LL k=0;												/*含质因数pi的数量*/ 
	for(LL i=n;i;i/=pi)k+=i/pi;
	for(LL i=m;i;i/=pi)k-=i/pi;
	for(LL i=n-m;i;i/=pi)k-=i/pi;
	return nn*inv(mm,p)*inv(nm,p)*ksm(pi,k,p)%p;
}
LL crt(LL len,LL Lcm)
{
	LL ans=0;
	for(LL i=1;i<=len;++i){
		LL Mi=Lcm/M[i];
		ans=((ans+A[i]*inv(Mi,M[i])*Mi)%Lcm+Lcm)%Lcm;
	}
	return ans;
}
int main()
{
	ios::sync_with_stdio(false);
	LL n,m,P,num;
	while(cin>>n>>m>>P){
		if(n<m){
			cout<<0<<endl;
			continue;
		}
		num=0;
		memset(A,0,sizeof(A));
		memset(M,0,sizeof(M));
		for(LL x=P,i=2;i<=P;++i)
			if(x%i==0){
				M[++num]=1;
				while(x%i==0){
					M[num]*=i;
					x/=i;
				}
				A[num]=exlucas(n,m,i,M[num])%P;
			} 
		cout<<crt(num,P)<<endl;
	}
	return 0;
}
## 十, 多项式算法 ### FFT(根据复数, 答案是小数, 非精确) ### NTT 
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#include <bits/stdc++.h>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 4000005, maxm = 100005, INF = 1000000000;

inline int read() {
    int out = 0, flag = 1; char c = getchar();
    while (c < 48 || c > 57) { if (c == '-') flag = -1; c = getchar(); }
    while (c >= 48 && c <= 57) { out = (out << 3) + (out << 1) + c - 48; c = getchar(); }
    return out * flag;
}

const int G = 3, P = (119 << 23) + 1;
int n, m, L, R[maxn];
int A[maxn], B[maxn];

int qpow(int a, int b) {
    int ans = 1;
    for (; b; b >>= 1, a = 1ll * a * a % P)
        if (b & 1) ans = 1ll * ans * a % P;
    return ans;
}

void NTT(int* a, int f) {
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i], a[R[i]]);
    for (int i = 1; i < n; i <<= 1) {
        int gn = qpow(G, (P - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)) {
            int g = 1;
            for (int k = 0; k < i; k++, g = 1ll * g * gn % P) {
                int x = a[j + k], y = 1ll * g * a[j + k + i] % P;
                a[j + k] = (x + y) % P; a[j + k + i] = (x - y + P) % P;
            }
        }
    }
    if (f == 1) return;
    int nv = qpow(n, P - 2); reverse(a + 1, a + n);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}

int main() {
    n = read();
    m = read();
    for (int i = 0; i <= n; i++) A[i] = read();
    for (int i = 0; i <= m; i++) B[i] = read();
    m = n + m;                                //m = n + m!
    for (n = 1; n <= m; n <<= 1)
        L++;
    for (int i = 0; i < n; i++)
        R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    NTT(A, 1);
    NTT(B, 1);
    for (int i = 0; i < n; i++)
        A[i] = 1ll * A[i] * B[i] % P;
    NTT(A, -1);
    for (int i = 0; i <= m; i++)
        printf("%d ", A[i]);
    return 0;
}
## 十一, 康托展开/逆康托展开 ### 康托展开, 某排列映射一个值 
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//对前 10 个自然数(0 ~ 9)的阶乘存入表
//以免去对其额外的计算
const int fact[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
/**
 * @brief 康拓展开
 *
 * @param[in] permutation 输入的一个全排列
 * @param[out] num 输入的康拓映射，即是第几个全排列
 */
int contor(const vector<int>& permutation) {
    int num = 0;
    int len = permutation.size();
    for (int i = 0; i < len; ++i) {
        int cnt = 0; // 在 i 之后，比 i 还小的有几个
        for (int j = i + 1; j < len; ++j)
            if (permutation[i] > permutation[j]) ++cnt;
        num += cnt * fact[len - i - 1];
    }
    return num + 1;
}
### 反求排列 
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//对前 10 个自然数(0 ~ 9)的阶乘存入表
//以免去对其额外的计算
const int fact[10] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};
/**
 * @brief 逆康拓展开
 *
 * @param[in] bits 给定全排列的使用数字个数
 * @param[in] num 给定全排列的次位
 * @param[out] permutation 输出对应的全排列
 */
vector<int> revContor(int bits, int num) {
    num = num - 1; //有 num - 1 个排列比目标序列要小
    vector<bool> vis(bits + 1, false);
    vector<int> permutation(bits, -1);

    int n, residue = num;
    for (int i = 0; i < bits; ++i) {
        n = residue / (fact[bits - i - 1]);
        residue = residue % (fact[bits - i - 1]);

        for (int j = 1; j <= bits; ++j) {
            if (!vis[j] && !(n--)) {
                vis[j] = true;
                permutation[i] = j;
                break;
            }
        }
    }
    return permutation;
}
## 十二, 日期计算 ### 吉姆拉尔森公式 
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int CaculateWeekDay(int y, int m, int d) { //年月日, 注意: 1/2月要当成上一年的13/14月
    if (m == 1 || m == 2) {
        m += 12;
        y--;
    }
    int iWeek = (d + 2 * m + 3 * (m + 1) / 5 + y + y / 4 - y / 100 + y / 400) % 7;
    return iWeek; // (0 - 6)
}
## 十三, 容斥原理 递归/暴力 若在n较大的情况下，应该采用递归进行计算，位运算计算循环次数太多，反而很慢。 在集合1，2，3…600,中，取出能被2，3，5，整除的数的个数 ### 位操作: 
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#include<bits/stdc++.h>
using namespace std;

int p[] = { 2,3,5 };
int cnt = 3;

int cal(int n = 600) {
    int res = 0;
    for (int i = 1; i < (1 << cnt); i++) {
        int t = i, tmp = 1, k = 0;
        int len = 0;
        while (t) {
            if (t & 1) {
                tmp *= p[k];
                len++;
            }
            t >>= 1;
            k++;
        }
        if (len & 1) res += n / tmp;
        else res -= n / tmp;
    }
    return res;
}


int main() {
    cout << cal() << endl;
    return 0;
}
### 递归版 
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#include<bits/stdc++.h>
using namespace std;

int a[] = { 2,3,5 };

int b = 600;
int sum = 0;
int n = 3;


void dfs(int i, int num, int x, int mu) {  //i表示第几个元素，num表示现在一共用了几个，x表示最多能用几个，mu表示当前取了的数的乘积
    if (num == x) {
        sum += b / mu;                //b就表示600
        return;
    }

    if (i == n) return;             //一共有n个，比如上面的2，3，5，一共有3个，
    dfs(i + 1, num + 1, x, mu * a[i]);    //a中存储的就是2，3，5，然后取或者不取
    dfs(i + 1, num, x, mu);
}

int rong() {
    int s = 0;
    for (int i = 1; i <= n; i++) {
        sum = 0;
        dfs(0, 0, i, 1);
        if (i & 1) s += sum;        //容斥定理
        else s -= sum;
    }
    return s;                    //s为能被2，3，5整除的数的个数。
}


void dfs(int i, int mu, int num) {
    mu *= a[i];
    if (num % 2) sum += b / mu;
    else sum -= b / mu;
    for (int j = i + 1; j < 3; j++) {
        dfs(j, mu, num + 1);
    }
}

int main() {
    printf("%d\n", rong());
    sum = 0;
    for (int i = 0; i < 3; i++) {
        dfs(i, 1, 1);
    }
    printf("%d\n", sum);
    return 0;
}
### 从 m 中颜色中选择 k 种, 给 n 个盒子染色, 相邻盒子颜色不同 ### 打表求 组合数(取模) 用不超过 k 种颜色涂完 n 个盒子的方案数为 $ x\_k=k(k−1)^{n−1} $ n 个盒子，用 m 种颜色中选出 正好 k 种颜色，且相邻盒子颜色不同的方案数为: $$ ans = C\_{m}^{k} t m p=C\_{m}^{k} x\_{k}-C\_{m}^{k-1} x\_{k-1}+C\_{m}^{k-2} x\_{k-2}+\\ldots+(-1)^{k} x\_{1} $$ 
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#include<stdio.h>
#include<cstdio>
#include<iostream>
using namespace std;

#define ll long long

const ll mod=1e9+7;
const int maxn=1e6+10;

int T;
ll p[maxn], invp[maxn], inv[maxn];

//(a^b)%p
ll qp(ll a, ll b){
    a %= mod;
    ll ret = 1;
    while(b){
        if (b & 1)
            ret = ret * a % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ret;
}

void init(){
    p[0] = 1;
    for(int i = 1; i <= 1000000; i++)
        p[i] = p[i - 1] * i % mod;
    invp[1000000] = qp(p[1000000], mod - 2);
    for(int i = 1000000 - 1; i>=0; i--)
        invp[i] = invp[i + 1] * (i + 1) % mod;
    inv[1] = 1;
    for(int i = 2; i <= 1000000; i++)
        inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
}

ll C(ll n, ll m){
    return p[n] * invp[m] % mod * invp[n - m] % mod;
}

int main()
{
    init();
    scanf("%d",&T);
    while(T--)
    {
        ll n,m,k;
        scanf("%lld%lld%lld",&n,&m,&k);
        ll Cmk = 1;
        for(int i=1; i<=k; i++)
            Cmk = Cmk * (m - k + i) % mod * inv[i] % mod;
        ll tmp = 0;
        for(int i=0; i<k; i++)
            if (i % 2 == 0)
                tmp = (tmp + C(k, k-i) * (k - i) % mod * qp(k - i - 1, n - 1) % mod + mod) % mod;
            else
                tmp = (tmp - C(k, k-i) * (k - i) % mod * qp(k - i - 1, n - 1) % mod + mod) % mod;
        ll ans = Cmk * tmp % mod;
        printf("%lld\n", ans);
    }
    return 0;
}
## 十四, 欧拉降幂 输入n, m 长度n序列, m是mod q个查询, 每次输入l, r 求: $$ w\[l\]^{w\[l+…\]^{w\[r\]}}$$ 
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#include <bits/stdc++.h>
//#include <hash_map>
//#include <unordered_map> //没快多点
using ll = long long;
using namespace std;

#define MOD(a,b) a>=b?a%b+b:a  //重定义取模，按照欧拉定理的条件
const int maxn = 1e6 + 10;

int is_prime[maxn];
int prime[maxn];
int phi[maxn];
int cnt;

void init() {
    cnt = 0; //统计素数个数
    is_prime[1] = 1; //0代表是素数, 1不是素数
    phi[1] = 1;
    for (int i = 2; i <= maxn - 10; i++) {
        
        if (!is_prime[i]) {
            prime[++cnt] = i; //i是素数, prime从下标[1]开始有记录
            phi[i] = i - 1;
        }
        for (int j = 1; j <= cnt; j++) { // 1 到 cnt !
            if (prime[j] * i >= maxn)
                break;
            is_prime[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            else {
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);
            }
        }
    }
}

unordered_map<int, int> ma;
int n;
ll w[maxn], m, l, r, x;

ll qpow(ll a, ll b, ll mod) { //快速幂
    ll res = 1;
    while (b) {
        if (b & 1) res = MOD(res * a, mod);
        a = MOD(a * a, mod);
        b >>= 1;
    }
    return res;
}

ll getphi(ll x) {  //求互质个数
    if (x <= maxn - 10) {
        return phi[x];
    }
    ll ans = x;
    if (ma.count(x)) return ma[x];
    for (int i = 1; 1ll*prime[i] * prime[i] <= x; i++) {
        if (x % prime[i] == 0) {
            ans = ans / prime[i] * (prime[i] - 1);
            while (x % prime[i] == 0) {
                x /= prime[i];
            }
        }
        /*
        if (x % i == 0) {
            ans = ans / i * (i - 1);
            while (x % i == 0)
                x /= i;
        }
        */
    }
    if (x > 1) ans = ans / x * (x - 1); //*****
    ma[x] = ans;
    return ans;
}
ll solve(int ceng, ll mod) { //每层的幂是w[i]
    if (ceng == r || mod == 1)
        return MOD(w[ceng], mod);
    return
        qpow(w[ceng], solve(ceng + 1, getphi(mod)), mod);
}

int main() {
    init();
    //cin >> n >> m;
    scanf("%d %lld", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &w[i]);
    }
    int q;
    //cin >> q;
    scanf("%d", &q);
    while (q--) {
        //cin >> l >> r;
        scanf("%lld %lld", &l, &r);
        //r = rr;
        cout << solve(l, m) % m << '\n';
    }
    return 0;
}
## 十五, 高斯消元 
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const double EPS = 1E-9;
int n;
vector<vector<double> > a(n, vector<double>(n));

double det = 1;
for (int i = 0; i < n; ++i) {
  int k = i;
  for (int j = i + 1; j < n; ++j)
    if (abs(a[j][i]) > abs(a[k][i])) k = j;
  if (abs(a[k][i]) < EPS) {
    det = 0;
    break;
  }
  swap(a[i], a[k]);
  if (i != k) det = -det;
  det *= a[i][i];
  for (int j = i + 1; j < n; ++j) a[i][j] /= a[i][i];
  for (int j = 0; j < n; ++j)
    if (j != i && abs(a[j][i]) > EPS)
      for (int k = i + 1; k < n; ++k) a[j][k] -= a[i][k] * a[j][i];
}

cout << det;
## 十六, 杜教BM 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long long ll;
typedef vector<int> VI;
const int maxn = 10005;
const ll mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
ll fast_mod(ll a, ll n, ll Mod)
{
    ll ans = 1;
    a %= Mod;
    while (n)
    {
        if (n & 1)
            ans = (ans * a) % Mod;
        a = (a * a) % Mod;
        n>>= 1;
    }
    return ans;
}
namespace linear_seq
{
ll res[maxn], base[maxn], num[maxn], md[maxn]; // 数组大小约 10000
vector<int> vec;
void mul(ll *a, ll *b, int k)
{
    for (int i = 0; i <2 * k; i++)
        num[i] = 0;
    for (int i = 0; i < k; i++)
    {
        if (a[i])
            for (int j = 0; j < k; j++)
                num[i + j] = (num[i + j] + a[i] * b[j]) % mod;
    }
    for (int i = 2 * k - 1; i>= k; i--)
    {
        if (num[i])
            for (int j = 0; j <vec.size(); j++)
                num[i - k + vec[j]] = (num[i - k + vec[j]] - num[i] * md[vec[j]]) % mod;
    }
    for (int i = 0; i < k; i++)
        a[i] = num[i];
}
ll solve(ll n, VI a, VI b)
{
    ll ans = 0, cnt = 0;
    int k = a.size();
    assert(a.size() == b.size());
    for (int i = 0; i < k; i++)
        md[k - 1 - i] = -a[i];
    md[k] = 1;
    vec.clear();
    for (int i = 0; i < k; i++)
        if (md[i])
            vec.push_back(i);
    for (int i = 0; i < k; i++)
        res[i] = base[i] = 0;
    res[0] = 1;
    while ((1LL << cnt) <= n)
        cnt++;
    for (int p = cnt; p>= 0; p--)
    {
        mul(res, res, k);
        if ((n>> p) & 1)
        {
            for (int i = k - 1; i>= 0; i--)
                res[i + 1] = res[i];
            res[0] = 0;
            for (int j = 0; j < vec.size(); j++)
                res[vec[j]] = (res[vec[j]] - res[k] * md[vec[j]]) % mod;
        }
    }
    for (int i = 0; i < k; i++)
        ans = (ans + res[i] * b[i]) % mod;
    if (ans < 0)
        ans += mod;
    return ans;
}
VI BM(VI s)
{
    VI B(1, 1), C(1, 1);
    int L = 0, m = 1, b = 1;
    for (int i = 0; i < s.size(); i++)
    {
        ll d = 0;
        for (int j = 0; j < L + 1; j++)
            d = (d + (ll)C[j] * s[i - j]) % mod;
        if (d == 0)
            m++;
        else if (2 * L <= i)
        {
            VI T = C;
            ll c = mod - d * fast_mod(b, mod - 2, mod) % mod;
            while (C.size() < B.size() + m)
                C.push_back(0);
            for (int j = 0; j < B.size(); j++)
                C[j + m] = (C[j + m] + c * B[j]) % mod;
            L = i + 1 - L, B = T, b = d, m = 1;
        }
        else
        {
            ll c = mod - d * fast_mod(b, mod - 2, mod) % mod;
            while (C.size() < B.size() + m)
                C.push_back(0);
            for (int j = 0; j < B.size(); j++)
                C[j + m] = (C[j + m] + c * B[j]) % mod;
            m++;
        }
    }
    return C;
}
int gao(VI a, ll n)
{
    VI c = BM(a);
    c.erase(c.begin());
    for (int i = 0; i < c.size(); i++)
        c[i] = (mod - c[i]) % mod;
    return solve(n, c, VI(a.begin(), a.begin() + c.size()));
}
} // namespace linear_seq
int main()
{
    //freopen("in.txt", "r", stdin);
    // 填数字的时候带上模数之后的
    ll t, n;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld", &n);
        // 在 vi 中填入自己推的数 8 个以上 越多越好
        printf("%lld\n", linear_seq::gao(VI{3, 9, 20, 46, 106, 244, 560, 1286, 2956, 6794}, n-1));
    }
    return 0;
}
## 十七, 拉格朗日插值 高数的应用: 根据导数, 控制函数大小 这里的应用: 给出n个点P(xi, yi)，将过这个点的最多n - 1次的多项式记为f(x)，求f(k)的值 
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#include <cstdio>

const int maxn = 2010;
using ll = long long;
ll mod = 998244353;
ll n, k, x[maxn], y[maxn], ans, s1, s2;

ll powmod(ll x, ll n) {
  ll ret = 1ll;
  while (n) {
    if (n & 1) ret = ret * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}

ll inv(ll x) { return powmod(x, mod - 2); }

int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 1; i <= n; i++)
    scanf("%lld%lld", x[i], y[i]);
  for (int i = 1; i <= n; i++) {
    s1 = y[i] % mod;
    s2 = 1ll;
    for (int j = 1; j <= n; j++)
      if (i != j) s1 = s1 * (k - x[j]) % mod, s2 = s2 * (x[i] - x[j]) % mod;
    ans += s1 * inv(s2) % mod;
  }
  printf("%lld\n", (ans % mod + mod) % mod);
  return 0;
}
## 十八, Catalan 数列 常常用来求解这样一类问题: 1. 有2n个人排成一行进入剧场。入场费5元。其中只有n个人有一张5元钞票，另外n人只有10元钞票，剧院无其它钞票，问有多少中方法使得只要有 10 元的人买票，售票处就有 5 元的钞票找零? 2. 在圆上选择2n个点，将这些点成对连接起来使得所得到的n条线段不相交的方法数？ 3. n个不同的数依次进栈，求不同的出栈结果的种数? 这一类, 有两种选择, 每种选择只能选择有限次数, 并且两种选择的可选择次数相同, 问多少种方案的问题. 
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#include <iostream>
using namespace std;
int n;
long long f[25];
int main() {
  f[0] = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) f[i] = f[i - 1] * (4 * i - 2) / (i + 1);
  // 这里用的是常见公式2
  cout << f[n] << endl;
  return 0;
}
## 十九, 常见的期望和方差 #### 0-1分布 1 次实验 P{x = 0} = p P{x = 1} = 1 - p E(x) = p; D(x) = pq; #### 二项分布 B(n, p) n次独立重复实验 P{x = k} = p E(x) = np; D(x) = npq; #### 几何分布 在n次独立重复实验中, 前k-1次皆失败，第k次成功的概率 P{x = k} = p E(x) = 1/p; D(x) = (1 - p) / (p \* p);