快速幂模版

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5,INF=0x3f3f3f3f;
typedef long long LL;
LL a,b,p; 
LL power(){
	LL ans=1;
	while(b){
		if(b&1)ans = ans*a%p;
		b>>=1;
		a = a*a%p;
	}
	return ans%p;
}
int main()
{
	cin>>a>>b>>p;
	cout<<power();
	return 0;
}

```
#include<iostream>
using namespace std;

int main()
{
    long long a, b, p, res = 1;
    scanf ("% ld % ld % ld", &a, &b, &p);
    while (b != 0)
   {
        if (b & 1)
        {
            res = res * a % p;
        }
        a = a * a % p;
        b >>= 1;
    }
    printf ("% ld \ n", res % p);
    return 0;
}
```

#include<bits/stdc++.h>
using namespace std;
int main(){
    long long a,b,p,res=1;
    scanf("%ld%ld%ld",&a,&b,&p);
    while(b!=0){
        if(b&1){
            res=res*a%p;
        }
        a=a*a%p;
        b>>=1;
    }
    printf("%ld\n",res%p);
    return 0;
}

知识点：快速幂

/*
int      %o/%lo(八进制) %d/%i/%ld/%li(十进制) %x/%lx(十六进制)[如标名为o/lo/d/i/lo/li/x/lx即输出为八进制/十进制/十六进制]
longlong %lld
float    %f/%e
double   %lf/%le
char     %c
char[]   %s
'a'=97
'z'=122
'A'=65
'Z'=90
'0'=48
'9'=57
*/
#include <iostream>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <limits>
#include <assert.h>
#include <stdlib.h>
using namespace std;
#define LL long long
#define ull unsigned long long
const int N=1e5+10;
const int INF=0x3f3f3f3f;
const double pi=3.1416;
LL a,b,p;
long long power(long long a,long long b,long long p){
	long long ans=1;
	long long wp=a;
	while(b){
		if(b&1){
			ans=ans*wp%p;
		}
		b>>=1;
		wp=wp*wp%p;
	}
	return ans%p;
}
int main(){
	cin>>a>>b>>p;
	cout<<power(a,b,p)<<endl;
return 0;
}

/*****************************************
备注：数学nanhai 5 
******************************************/
#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
LL power(LL n , LL m , LL p)
{
	if(n == 0)
		return 0;
	LL ans = 1;
	while(m)
	{
		if(m&1)
			ans = (ans * n) %p;
		m >>= 1;
		n = (n*n)%p;
	}
	return ans%p;
}
int main()
{
	LL n , m , p;
	cin >> n >> m >> p;
	cout << power(n , m , p);
	return 0;
}