I won't tell you this is about number theory （结论题）_to think of a beautiful problem description is so -优快云博客

本文介绍了一种快速解决特定数论问题的方法，该问题要求计算两个幂次表达式的最大公约数，并给出了一段C++代码实现。通过运用数学定理简化了计算过程。

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I won’t tell you this is about number theory

 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

题意：

求S

$S = gcd(a^m-1,a^n-1)%k$

分析：

有结论fuck！

设 $a > b,gcd(a,b) = 1$ 则 $gcd(a^m-b^m,a^n-b^n) = a^{gcd(m,n)} - b^{gcd(m,n)}$

特别的有 $a > 1,m,n > 0$ ，则 $gcd(a^m-1,a^n-1) = a^{gcd(m,n)}-1$

code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}
ll q_pow(ll a,ll b,ll mod){
    ll ans = 1;
    while(b){
        if(b & 1)
            ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        ll a,m,n,k;
        scanf("%lld%lld%lld%lld",&a,&m,&n,&k);
        ll g = gcd(m,n);
        printf("%lld\n",(q_pow(a,g,k) - 1 + k) % k);
    }
    return 0;
}