POJ 2154 color 波利亚群置换

 

Color

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 4071 Accepted: 1371

Description

Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of the necklace can be produced. You should know that the necklace might not use up all the N colors, and the repetitions that are produced by rotation around the center of the circular necklace are all neglected.

You only need to output the answer module a given number P.

Input

The first line of the input is an integer X (X <= 3500) representing the number of test cases. The following X lines each contains two numbers N and P (1 <= N <= 1000000000, 1 <= P <= 30000), representing a test case.

Output

For each test case, output one line containing the answer.

Sample Input

51 300002 300003 300004 300005 30000

Sample Output

131170629

Source

POJ Monthly,Lou Tiancheng

 

解法就是用波利亚原理

 

不知道为什么，我用__int64就会超时，但是用int就会WA

我考虑到可能是二分幂的时候会超过int，于是换了很多个写法的二分幂模板。最后只有一个可以过

 

我的代码：

 

#include<stdio.h>#include<string.h>typedef int ll;ll prime[35000],m;bool flag[35000];ll eular(ll n){ll ret=1,i;for(i=2;i*i<=n;i++){if(n%i==0){ret=ret*(i-1);n=n/i;while(n%i==0){n=n/i;ret=ret*i;}}if(n==1)break;}if(n>1)ret=ret*(n-1);return ret%m;}ll exmod(ll p,ll n){     ll sq=1; while(n>0) { if(n%2==1) sq=sq*p%m; p=(p%m)*(p%m)%m; n/=2; } return sq%m;}int main(){ll n,t,ans,i;scanf("%d",&t);while(t--){ans=0;scanf("%d%d",&n,&m);for(i=1;i*i<n;i++){if(n%i==0){ans=(ans+eular(n/i)*exmod(n,i-1)%m)%m;ans=(ans+eular(i)*exmod(n,n/i-1)%m)%m;}}if(i*i==n)ans=(ans+eular(i)*exmod(n,i-1)%m)%m;printf("%d\n",ans%m);}return 0;}