【POJ 2154】Color(置换群)
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Color
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 9265 Accepted: 3010
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
[题意][将正n边形的n个顶点用n种颜色染色,问有多少种方案(答案mod p,且可由旋转互相得到的算一种)]
【题解】【与前面几题相似,但此题没有旋转,但数据范围相当恶心,所以,要优化】
=
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按照这个公式筛欧拉函数,但因为范围过大,所以,不可能全部筛出,筛一半,另一半,用下面的方法
根号n的时间内求一个数的phiinline int find(int m){ int i,k=m,l=m; for (i=2;i*i<=m;++i) if (!(l%i)) { k=k-k/i; do{ l/=i; }while(!(l%i)); } if (l>1) k=k-k/l; return k;}
由于范围过大,枚举的时候只能枚举一半,而且要谨慎使用long long
#include<cstdio>#include<cstring>#include<algorithm>#define ll long longusing namespace std;int prime[100010],phi[100010];int m,n,mod;bool p[100010];inline void shai(int m){ int i,j; phi[1]=1; p[1]=1; for(i=2;i<=m;++i) { if(!p[i]) prime[++prime[0]]=i,phi[i]=i-1; for(j=1;j<=prime[0];++j) { if(i*prime[j]>m) break; p[i*prime[j]]=1; if(!(i%prime[j])) {phi[i*prime[j]]=phi[i]*prime[j]; break;} else phi[i*prime[j]]=phi[i]*(prime[j]-1); } } return;}ll poww(int x,int q){ if(!q) return 1; if(q==1) return x%mod; if(q==2) return x*x%mod; if(q%2) {ll sum=poww(x,q/2)%mod; sum=sum*sum*x%mod; return sum%mod;} else {ll sum=poww(x,q/2)%mod; sum*=sum; return sum%mod; }}inline ll get_phi(int x){ if(x<=70000) return phi[x]; ll ans=x; int i; for(i=2;i*i<=x;++i) if(!(x%i)) { ans=ans*(i-1)/i; while(!(x%i)) x/=i; } if(x>1) ans=ans*(x-1)/x; return ans%mod;}int main(){ int i,j; shai(100000); scanf("%d",&n); for(j=1;j<=n;++j) { ll ans=0; scanf("%d%d",&m,&mod); for(i=1;i*i<=m;++i) if(!(m%i)) { ll s1=get_phi(i); ll s2=get_phi(m/i); if(m/i==i) ans=(ans+s2*poww(m%mod,i-1))%mod; else ans=(ans+s2*poww(m%mod,i-1))%mod, ans=(ans+s1*poww(m%mod,m/i-1))%mod; } printf("%I64d\n",ans%mod); } return 0;}
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