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[ACM] POJ 2154 Color (Polya计数优化,欧拉函数)

时间:2014-07-31 00:10:35      阅读:330      评论:0      收藏:0      [点我收藏+]

标签:acm   组合   

Color
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 7630   Accepted: 2507

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


解题思路:

转载:http://blog.csdn.net/tsaid/article/details/7366708

题意:给出两个整数n和p,代表n个珠子,n种颜色,要求不同的项链数,翻转置换不考虑。结果模p.

题解:

我们知道gcd(i,n)表示了循环节的个数。例如gcd(2,6) = 2, 它的具体过程为:[1,3,5] [2,4,6]

对于任意一个循环置换,他所有循环节的长度为 n / gcd(i,n),在上面的例子中: 循环节长度 = 6 / gcd(2,6) = 3


为了方便说明,用L表示循环节的长度,显然 L | n

如果我们枚举L,求出对于每一个L有多少个i, 使得 L = n / gcd (i,n), 那么我们实际上也得到了循环节个数为 n / L 的置换个数。


将L = n / gcd (i,n)转换一下得到:n / L = gcd(i,n )

设 cnt = n / L = gcd(i, n)  注:cnt表示循环节个数,L表示每一个循环节的长度

因为 cnt | i, 所以可令 i = cnt * t; ( 因为0 <= i < n, 所以0 <= t < n / cnt = L )

又因为 cnt = n / L, 所以 n = cnt * L;

则 gcd ( i, n ) = gcd ( cnt*t, cnt*L ) = cnt;  ①

可知当 gcd ( t, L ) = 1 时 ① 式成立。

由于 gcd ( t, L ) = 1 的个数就是 Euler(L)的个数,

所以我们可以得到结论:循环节个数为n/L的置换有Euler(L)个

代码:

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
bool isprime[50001];
int prime[50001];
int len=0;;
int n,p;

void sieve()
{
    for(int i=0;i<=50000;i++)
        isprime[i]=1;
    isprime[0]=isprime[1]=0;
    for(int i=2;i<=50000;i++)
    {
        if(isprime[i])
        {
            prime[len++]=i;
            for(int j=2*i;j<=50000;j+=i)
                isprime[j]=0;
        }
    }
}

int euler(int n)
{
    int res=n;
    for(int i=0;i<len&&prime[i]*prime[i]<=n;i++)
    {
        if(n%prime[i]==0)
        {
            res=res/prime[i]*(prime[i]-1);
            while(n%prime[i]==0)
                n/=prime[i];
        }
    }
    if(n>1)
        res=res/n*(n-1);
    return res;
}

int pow(int p,int n,int mod)
{
    int ans=1;
    p=p%mod;
    while(n)
    {
        if(n&1)
            ans=ans*p%mod;
        p=p*p%mod;
        n/=2;
    }
    return ans;
}

int main()
{
    sieve();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int ans=0;
        scanf("%d%d",&n,&p);
        for(int i=1;i*i<=n;i++)
            if(n%i==0)
            {
                ans=(ans+euler(i)%p*pow(n,n/i-1,p))%p;//注意取余的位置, euler(i)%p不取余就WA
                if(i*i==n)//只要一个i就可以了
                    break;
                ans=(ans+euler(n/i)%p*pow(n,i-1,p))%p;
            }
        printf("%d\n",ans);
    }
    return 0;
}



[ACM] POJ 2154 Color (Polya计数优化,欧拉函数),布布扣,bubuko.com

[ACM] POJ 2154 Color (Polya计数优化,欧拉函数)

标签:acm   组合   

原文地址:http://blog.csdn.net/sr_19930829/article/details/38306469

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