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+];

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

 LL Euler (LL n) {
     LL rea=n;
     for(int i=; i*i<=n; i++)
         if(n%i==)
         {
             rea=rea-rea/i;
             do
                 n/=i;
             while(n%i==);
         }
     if(n>)
         rea=rea-rea/n;
     return rea;
 }
 LL jieguo(LL a,char b[],LL c,LL mod) {
     LL sum = ;
     sum = (b[] - '') % mod;
     LL len = strlen(b);
     for(int i = ;i<len;i++) {
         sum*=;
         sum=(sum+b[i]-'')%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\n",jieguo(c,b,d,ola));
     }
     return ;
 }