【模板】欧拉定理

题目背景

模板题,无背景

题目描述

给你三个正整数,a,m,ba,m,ba,m,b,你需要求:
abmodma^b mod mabmodm

输入输出格式

输入格式:

一行三个整数,a,m,ba,m,ba,m,b

输出格式:

一个整数表示答案

输入输出样例

输入样例#1: 复制
2 7 4
输出样例#1: 复制
2
输入样例#2: 复制
998244353 12345 98765472103312450233333333333
输出样例#2: 复制
5333

说明

注意输入格式,a,m,ba,m,ba,m,b 依次代表的是底数、模数和次数

样例1解释:
24mod7=22^4 mod 7 = 224mod7=2
输出2

数据范围:
对于全部数据:
1≤a≤1091≤a≤10^91a109
1≤b≤10200000001≤b≤10^{20000000}1b1020000000
1≤m≤1061≤m≤10^61m106

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 98765431;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int a, m, phi = 1;
int B, fg;
ll qpow(ll x, ll y) {
	ll ans = 1;
	while (y) {
		if (y % 2)ans = ans * x%m;
		x = x * x%m; y >>= 1;
	}
	return ans;
}

int main()
{
	//	ios::sync_with_stdio(0);
	rdint(a); rdint(m);
	a %= m; int tmp = m;
	for (int i = 2; i <= sqrt(tmp); i++) {
		if (tmp%i)continue;
		phi *= (i - 1); tmp /= i;
		while (tmp%i == 0) {
			phi *= i; tmp /= i;
		}
	}
	if (tmp > 1)phi *= (tmp - 1);
	char ch;
	while ((ch = getchar()) < '0' || ch > '9');
	while (B = B * 10ll + (ch^'0'), (ch = getchar()) >= '0'&&ch <= '9') {
		if (B >= phi)fg = 1, B %= phi;
	}
	if (B >= phi)fg = 1, B %= phi;
	if (fg)B += phi;
	printf("%lld
", qpow(a * 1ll, B * 1ll));
	return 0;
}
原文地址:https://www.cnblogs.com/zxyqzy/p/10372861.html