POJ2154 Color

Posted 2020-09-09 SilverNebula

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

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 #define LL long long
 9 using namespace std;
10 const int mxn=100101;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
15     return x*f;
16 }
17 int pri[mxn],cnt=0;
18 bool vis[mxn];
19 void init(){
20     for(int i=2;i<mxn;i++){
21         if(!vis[i])
22             pri[++cnt]=i;
23         for(int j=1;j<=cnt && (long long)pri[j]*i<mxn;j++){
24             vis[pri[j]*i]=1;
25             if(i%pri[j]==0)break;
26         }
27     }
28     return;
29 }
30 int phi(int x){
31     int res=x;
32     int m=sqrt(x+0.5);
33 //    for(int i=1;i<=cnt && x>1;i++){
34     for(int i=1;i<=cnt && pri[i]<=m;i++){
35         if(x%pri[i]==0){
36             res=res/pri[i]*(pri[i]-1);
37             while(x%pri[i]==0)
38                 x/=pri[i];
39         }
40     }
41 //    printf("x:%d %d\n",x,res);
42     if(x>1)res=res/x*(x-1);
43     return res;
44 }
45 int n,p;
46 int ksm(int c,int k){
47     int res=1;
48     while(k){
49         if(k&1)(res*=c)%=p;
50         c=c*c%p;
51         k>>=1;
52     }
53     return res;
54 }
55 int main(){
56     int i,j;
57     int T=read();
58     init();
59     while(T--){
60         n=read();p=read();
61         int ans=0;
62         for(i=1;i*i<n;i++){
63             if(n%i==0){
64 //                printf("i:%d phi:%d\n",n/i,phi(n/i));
65                 ans+=ksm(n%p,i-1)*(phi(n/i)%p);
66                 ans%=p;
67                 ans+=ksm(n%p,n/i-1)*(phi(i)%p);
68                 ans%=p;
69             }
70         }
71         if(i*i==n)ans=(ans+ksm(n%p,i-1)*(phi(i)%p))%p;
72         printf("%d\n",ans%p);
73     }
74     return 0;
75 }