hdu 2685(数论相关定理+欧几里德定理+快速取模)

I won't tell you this is about number theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 458    Accepted Submission(s): 142

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

 

Author

Teddy

 

Source

ZJFC 2009-3 Programming Contest

 

这道题要知道这个公式：

gcd(a^m-1,aⁿ-1) = a^gcd(m,n)-1

推广:

若 gcd(a,b)=1

gcd(a^m-b^m,aⁿ-bⁿ) = a^gcd(m,n)-b^gcd(m,n)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long LL;
LL pow_mod(LL a,LL n,LL mod){
    LL ans = ;
    while(n){
        if(n&) ans = ans*a%mod;
        a=a*a%mod;
        n=n>>;
    }
    return ans;
}
LL gcd(LL a,LL b){
    return b==?a:gcd(b,a%b);
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        LL a,m,n,k;
        scanf("%lld%lld%lld%lld",&a,&m,&n,&k);
        LL t = gcd(m,n);
        LL ans = (pow_mod(a,t,k)-+k)%k;
        printf("%lld\n",ans);
    }
    return ;
}