POJ2154 Color

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10322		Accepted: 3360

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

5
1 30000
2 30000
3 30000
4 30000
5 30000

Sample Output

1
3
11
70
629

Source

POJ Monthly,Lou Tiancheng

数学问题 统计 polya原理

仍然是项链的套路，没有翻转只有旋转同构。

需要用到欧拉函数优化

1、数据范围大，要先打好素数筛

2、for统计答案的时候要一起算n和n/i，这样只用循环到sqrt(n)

没加这两个T飞了

 /*by SilverN*/
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<vector>
 #define LL long long
 using namespace std;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int pri[mxn],cnt=;
 bool vis[mxn];
 void init(){
     for(int i=;i<mxn;i++){
         if(!vis[i])
             pri[++cnt]=i;
         for(int j=;j<=cnt && (long long)pri[j]*i<mxn;j++){
             vis[pri[j]*i]=;
             if(i%pri[j]==)break;
         }
     }
     return;
 }
 int phi(int x){
     int res=x;
     int m=sqrt(x+0.5);
 //    for(int i=1;i<=cnt && x>1;i++){
     for(int i=;i<=cnt && pri[i]<=m;i++){
         if(x%pri[i]==){
             res=res/pri[i]*(pri[i]-);
             while(x%pri[i]==)
                 x/=pri[i];
         }
     }
 //    printf("x:%d %d\n",x,res);
     if(x>)res=res/x*(x-);
     return res;
 }
 int n,p;
 int ksm(int c,int k){
     int res=;
     while(k){
         if(k&)(res*=c)%=p;
         c=c*c%p;
         k>>=;
     }
     return res;
 }
 int main(){
     int i,j;
     int T=read();
     init();
     while(T--){
         n=read();p=read();
         int ans=;
         for(i=;i*i<n;i++){
             if(n%i==){
 //                printf("i:%d phi:%d\n",n/i,phi(n/i));
                 ans+=ksm(n%p,i-)*(phi(n/i)%p);
                 ans%=p;
                 ans+=ksm(n%p,n/i-)*(phi(i)%p);
                 ans%=p;
             }
         }
         if(i*i==n)ans=(ans+ksm(n%p,i-)*(phi(i)%p))%p;
         printf("%d\n",ans%p);
     }
     return ;
 }