SCOJ 4423: Necklace polya

4423: Necklace

题目连接：

http://acm.scu.edu.cn/soj/problem.action?id=4423

Description

baihacker bought a necklace for his wife on their wedding anniversary.

A necklace with N pearls can be treated as a circle with N points where the

distance between any two adjacent points is the same. His wife wants to color

every point, but there are at most 2 kinds of color. How many different ways

to color the necklace. Two ways are said to be the same iff we rotate one

and obtain the other.

Input

The first line is an integer T that stands for the number of test cases.

Then T line follow and each line is a test case consisted of an integer N.

Constraints:

T is in the range of [0, 4000]

N is in the range of [1, 1000000000]

N is in the range of [1, 1000000], for at least 75% cases.

Output

For each case output the answer modulo 1000000007 in a single line.

Sample Input

6

1

2

3

4

5

20

Sample Output

2

3

4

6

8

52488

Hint

题意

有一个长度为n的环，环上每个点的颜色有两种，然后你可以旋转

问你本质不同的串有多少种

题解：

ploya裸题

答案ans[i] = 1/n sigma(d|n)phi(d)*2^(n/d)

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
long long quickpow(long long  m,long long n,long long k)//返回m^n%k
{
    long long b = 1;
    while (n > 0)
    {
          if (n & 1)
             b = (b*m)%k;
          n = n >> 1 ;
          m = (m*m)%k;
    }
    return b;
}
long long phi(long long n)
{
    long long tmp=n;
    for(long long i=2;i*i<=n;i++)
        if(n%i==0)
        {
            tmp/=i;tmp*=i-1;
            while(n%i==0)n/=i;
        }
    if(n!=1)tmp/=n,tmp*=n-1;
    return tmp;
}
void solve()
{
    int n;
    scanf("%d",&n);
    long long ans = 0;
    for(int i=1;i*i<=n;i++)
    {
        if(n%i==0)
        {
            ans = ans + phi(i)*quickpow(2,(n/i),mod)%mod*quickpow(n,mod-2,mod)%mod;
            ans%= mod;
            if(i!=n/i)
            {
                ans = ans + phi(n/i)*quickpow(2,i,mod)%mod*quickpow(n,mod-2,mod)%mod;
                ans%= mod;
            }
        }
    }
    cout<<ans<<endl;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)solve();
}