FZU 1759-Super A^B mod C

传送门：http://acm.fzu.edu.cn/problem.php?pid=1759

Accept: 1161    Submit: 3892
Time Limit: 1000 mSec    Memory Limit : 32768 KB

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 4

2 10 1000

Sample Output

1 24

 

 

题意：

给出a,b,c，求a的b次方对c取模的结果

 

思路：

题意好理解，但是主要是a，b，c的范围比较大，所以这里面需要一个降幂公式：

 

#include<stdio.h>
#include<string.h>
typedef long long LL;

const int MAXX = 1e6;
LL c,d;
char b[MAXX+10];

LL quick_pow(LL a, LL b, LL mod) {
    LL ans = 1;
    while(b > 0) {
        if(b&1) {
            b--;
            ans = ans *a %mod;
        }
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}

LL Euler (LL n) {  
    LL rea=n;  
    for(int i=2; i*i<=n; i++)  
        if(n%i==0)
        {  
            rea=rea-rea/i;  
            do  
                n/=i;
            while(n%i==0);  
        }  
    if(n>1)  
        rea=rea-rea/n;  
    return rea;  
}
LL jieguo(LL a,char b[],LL c,LL mod) {
    LL sum = 0;
    sum = (b[0] - '0') % mod;
    LL len = strlen(b);
    for(int i = 1;i<len;i++) {
        sum*=10;
        sum=(sum+b[i]-'0')%mod;
    }
    return quick_pow(a,sum+mod,c);
}
int main() {

    
    while(scanf("%lld%s%lld", &c, b, &d)!=EOF){
        LL ola=Euler(d);
        printf("%lld
",jieguo(c,b,d,ola));
    }
    return 0;
}