hdu 5868 Polya计数

Different Circle Permutation

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 218 Accepted Submission(s): 106


[align=left]Problem Description[/align]
You may not know this but it's a fact that Xinghai Square is Asia's largest city square. It is located in Dalian and, of course, a landmark of the city. It's an ideal place for outing any time of the year. And now:

There are N children from a nearby primary school flying kites with a teacher. When they have a rest at noon, part of them (maybe none) sit around the circle flower beds. The angle between any two of them relative to the center of the circle is always a multiple of 2πN but always not 2πN.

Now, the teacher raises a question: How many different ways there are to arrange students sitting around the flower beds according to the rule stated above. To simplify the problem, every student is seen as the same. And to make the answer looks not so great, the teacher adds another specification: two ways are considered the same if they coincide after rotating.

[align=left]Input[/align]
There are T tests (T≤50). Each test contains one integer N. 1≤N≤1000000000 (109). Process till the end of input.

[align=left]Output[/align]
For each test, output the answer mod 1000000007 (109+7) in one line.

[align=left]Sample Input[/align]

4
7
10

[align=left]Sample Output[/align]

3
5
15

[align=left]Source[/align]
2016 ACM/ICPC Asia Regional Dalian Online

/*
hdu 5868 Polya计数

problem:
给你n个人,围绕成圆坐下,任意两人之间的距离必需是2pi/n的倍数.求旋转等效的情况下有多少种方案数

solve:
相当于给你n个间距为2pi/n的点,然后进行黑白染色,黑点不能相邻. (黑点表示坐人)
考虑Polya计数的话,需要枚举长度且得到长度为i的方案数.
找规律可以发现 f
= f[n-1] + f[n-2],用矩阵快速幂可以快速求出

hhh-2016-09-20 19:26:01
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 200100;
const int inf = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
struct Matrix
{
ll ma[2][2];
Matrix()
{
memset(ma,0,sizeof(ma));
}
};

Matrix mat;
Matrix from;

Matrix Mul(Matrix a,Matrix b)
{
Matrix c;
for(int i = 0; i < 2; i++)
{
for(int j = 0; j < 2; j++)
{
c.ma[i][j] = 0;
for(int k = 0; k < 2; k++)
{
c.ma[i][j] = c.ma[i][j] + a.ma[i][k]*b.ma[k][j] % mod;
c.ma[i][j] %= mod;
}
}
}
return c;
}

Matrix Pow(int n)
{
Matrix cnt;
Matrix t = mat;
memset(cnt.ma,0,sizeof(cnt.ma));
for(int i = 0; i < 2; i++)
cnt.ma[i][i] = 1;
while(n)
{
if(n & 1)
cnt = Mul(cnt,t);
t = Mul(t,t);
n >>= 1;
}
return cnt;
}

void init()
{
mat.ma[0][0] = 1,mat.ma[0][1] = 1,mat.ma[1][0] = 1,mat.ma[1][1] = 0;
from.ma[0][0] = 3,from.ma[1][0] = 1,from.ma[0][1] = 0,from.ma[1][1] = 0;
}
int n;
ll f(int i)
{
if(i == 1)
return 1;
if(i == 2)
return 3;
Matrix t = Mul(Pow(i-2),from);
return t.ma[0][0];
}

ll pow_mod(ll a,ll n)
{
ll ret = 1;
a %= mod;
while(n)
{
if(n & 1) ret = ret*a%mod;
a = a*a%mod;
n >>= 1;
}
return ret%mod;
}

ll euler(ll n)
{
ll ans = n;
ll i;
for (i = 2; i*i <= n; i++)
{
if (n%i == 0)
{
while (n%i == 0)
n /= i;
ans = ans/i*(i-1) ;
}
}
if (n != 1)
ans = ans/n*(n-1);
return ans;
}

void cal(int n)
{
if(n == 1)
{
printf("2\n");
return ;
}
ll ans = 0;
for(int i = 1; i*i <= n; i++)
{
if(n % i == 0)
{
ans = (ans + f(i)*euler(n/i)%mod)%mod;
if( i*i != n)
{
ans = (ans + f(n/i)*euler(i)%mod)%mod;
}
}
}
//    cout <<ans <<endl;
ans = ans*pow_mod(n,mod-2)%mod;
printf("%I64d\n",ans);
}

int main()
{
init();
//    for(int i =1 ;i <= 10;i++)
//        cout << f(i) <<endl;
//    for(int i =1 ;i <= 10;i++)
//        cout << euler(i) <<endl;
while(scanf("%d",&n) != EOF)
{
init();
cal(n);
}
}