POJ 2154 Color (Polya定理&欧拉函数)

来源：互联网发布：python mysql的参数化编辑：程序博客网时间：2024/06/03 16:42

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

题意：用n种颜色涂环形的n个珠子，重复只考虑旋转，不考虑翻转，问能构成多少种？

题解：由Polya定理可以给出方案数：,有很多i与n的gcd值相同，所以优化，把指数相同的一起算

令k=gcd(i,n),求有多少个i使之成立，枚举k，然后求i的个数：

显然n是k的倍数，令i=k*t,n=k*s;所以gcd(t,s)=1;求多少i变成了求多少t，再看范围，0<=i<n;所以0<=t<n/k;这里s=n/k;

问题就变成了求小于s与s互质的t有多少个答案就是eula(s)=eula(n/k),(欧拉函数)。

为了避免算1/n，把这个乘到和式里面，就是n的d-1次方。

#pragma comment(linker, "/STACK:10240000,10240000")#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<queue>#include<set>#include<vector>#include<map>#include<stack>#include<cmath>#include<algorithm>using namespace std;const double R=0.5772156649015328606065120900;const int N=31630+5;//const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eps=1e-8;const double pi=acos(-1.0);typedef long long ll;ll mod,tot=0;bool vis[N];int pri[N];ll Pow(ll a,ll n){    a%=mod;    ll ans=1;    while(n)    {        if(n&1)            ans=ans*a%mod;        a=a*a%mod;        n>>=1;    }    return ans;}void init(){    memset(vis,false,sizeof vis);    for(int i=2;i<N;i++)    if(!vis[i])    {        pri[tot++]=i;        for(int j=i;j<N;j+=i)            vis[j]=true;    }}int cnt[100];ll a[100],all;void resolve(ll n)//素因子分解{    all=0;    memset(cnt,0,sizeof cnt);    for(int i=0;i<tot;i++)    if(n%pri[i]==0)    {        a[all++]=pri[i];        while(n%pri[i]==0) n/=pri[i],cnt[all-1]++;    }    if(n>1)    {        a[all++]=n;        cnt[all-1]++;    }}ll eula(ll n){    ll ans=n;    for(int i=0;i<all;i++)//枚举出来的数的因子肯定是n有的，直接用已有的因子那选    if(n%a[i]==0)    {        ans=ans/a[i]*(a[i]-1);    }    return ans;}ll dfs(int pos,ll mul,ll n)//dfs枚举所有因子,第pos个素因子，已经枚举的书为mul,原始的n{    if(mul>n/mul) return 0;//大于sqrt(n)的已经算了    ll ans=0;    if(mul==n/mul)//注意完全平方数    {        ans=(ans+eula(mul)*Pow(n,n/mul-1))%mod;        return ans;    }    if(pos==all)    {        ans=(ans+eula(mul)*Pow(n,n/mul-1))%mod;//因子成对出现，sqrt(n)为界限        ans=(ans+eula(n/mul)*Pow(n,mul-1))%mod;        return ans;    }    ll tmp=1;    for(int i=0;i<=cnt[pos];i++)//枚举每个素因子选多少个，0就是不选    {        ans=(ans+dfs(pos+1,mul*tmp,n))%mod;        tmp*=a[pos];    }    return ans;}int main(){    init();    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        ll n;        scanf("%I64d%I64d",&n,&mod);        resolve(n);//分解n        printf("%I64d\n",dfs(0,1,n));    }    return 0;}/**/

0 0