hdu2685I won't tell you this is about number theory（快速幂加gcd）

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Problem Description 
To think of a beautiful problem description is so hard for me that let’s just drop them off. :) 
Given four integers a,m,n,k,and S = gcd(a^m-1,a^n-1)%k,calculate the S.

Input 
The first line contain a t,then t cases followed. 
Each case contain four integers a,m,n,k(1<=a,m,n,k<=10000).

Output 
One line with a integer S.

Sample Input 
1 

1 1 1 1

Sample Output 
0
这里用到啦数学的一个定理

gcd(a^m-b^m, a^n-b^n) = a^gcd(m,n)-b^gcd(m,n); 
这里b=1;

这里用一个快速幂 取模的板子就ok啦

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string.h>
#include<iterator>
#include<vector>
#include<stdlib.h>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<sstream>
#define lowbit(x) (x&(-x))
typedef long long ll;
using namespace std;
int MOD;
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
long long pow_m(long long a,long long n)
{
    long long ret = 1;
    long long tmp = a%MOD;
    while(n)
    {
        if(n&1)ret = (ret*tmp)%MOD;
        tmp = tmp*tmp%MOD;
        n >>= 1;
    }
    return ret;
}
int main()
{
    int a,b,c;
    while(t--)
    {
        ll a,b,n,k;
        scanf("%lld%lld%lld%lld",&a,&m,&n,&k);
        MOD = k;
        ll p = gcd(m,n);
        printf("%lld\n",(MOD+pow_m(a,p)-1)%MOD);
    }
    return 0;
}