| 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.
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 ",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/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 ",ans%p); 73 } 74 return 0; 75 }