FZU 1759 Super A^B mod C 指数循环节

Problem 1759 Super A^B mod C

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

思路：指数循环节，数据水，并没有B<C&&A,C不互质的；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e5+10,M=1e6+10,inf=1e9+10;
const int mod=1000000007;
#define MAXN 10000010
ll quickpow(ll x,ll y,ll z)
{
    ll ans=1;
    while(y)
    {
        if(y&1)
        ans*=x,ans%=z;
        x*=x;
        x%=z;
        y>>=1;
    }
    return ans;
}
ll phi(ll n)
{
    ll i,rea=n;
    for(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;
}
char a[M];
int main()
{
    ll x,y,z,i,t;
    while(~scanf("%I64d%s%I64d",&x,a,&z))
    {
        t=strlen(a);
        ll p=phi(z);
        ll ans=0;
        for(i=0;i<t;i++)
        ans=(ans*10+a[i]-'0')%p;
        ans+=p;
        printf("%I64d\n",quickpow(x,ans,z));
    }
    return 0;
}