FZU1759Super A^B mod C(快速幂取模) 公式

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

对于A^B%C 有一个公式 即

A^x = A^(x % Phi(C) + Phi(C)) (mod C)

公式的具体证明：http://hi.baidu.com/aekdycoin/item/e493adc9a7c0870bad092fd9

// A^x = A^(x % Phi(C) + Phi(C)) (mod C)，其中x≥Phi(C)
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long  LL;
char bb[1000001];
LL a,m;
LL phi(LL n)
{
LL rea=n;
for(int i=2;i*i<=n;i++){
if(n%i==0){
rea=rea-rea/i;
while(n%i==0)
n/=i;
}
}
if(n>1)
rea=rea-rea/n;
return rea;
}
LL quick_mod(LL a,LL b,LL m)
{
LL ans=1;
a%=m;
while(b){
if(b&1){
ans=ans*a%m;
b--;
}
b>>=1;
a=a*a%m;
}
return ans;
}
int main()
{
while(~scanf("%lld",&a)){
scanf("%s",bb);
scanf("%lld",&m);
LL t=phi(m);
int l=strlen(bb);
LL b=0;
for(int i=0;i<l;i++){
b=b*10+bb[i]-'0';
while(b>=t)
b-=t;
}
b+=t;
printf("%lld\n",quick_mod(a,b,m));
}
return 0;
}