Problem 1759 Super A^B mod C (超高次幂）

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Problem Description

Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

Input

There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.

Output

For each testcase, output an integer, denotes the result of A^B mod C.

Sample Input

3 2 42 10 1000

Sample Output

124

Source

FZU 2009 Summer Training IV--Number Theory

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
 
const int maxn=1e6+10;
char B[maxn];
 
/*
欧拉降幂 
*/
ll A,C;
 
ll pow_(ll base,ll n){
	ll ans=1,mod=C;
	while(n){
		if(n&1){
			ans=ans*base%mod; 
		}
		base=base*base%mod;
		n>>=1;
	}
	return ans;
}
 
ll phi(ll x){
	ll ans=x;
	ll m=sqrt(x+0.5);
	for(ll i=2;i<=m;i++){
		if(x%i==0){
			ans=ans-ans/i;
		}
		while(x%i==0)x/=i;
	}
	if(x>1)ans=ans-ans/x;
	return ans;
}
 
int main()
{
	while(scanf("%I64d %s %I64d",&A,B,&C)==3){
	    ll cc=phi(C);
		int len=strlen(B);
		ll bb=0;

		//大整数 mod 
		for(int i=0;i<len;i++){
			bb=(bb*10+B[i]-'0')%cc; 
		}
		//bb+=cc;	

		ll ans=pow_(A,bb);
		printf("%I64d\n",ans);
	}
	return 0;
}