HDU 3923 Invoker（polya定理+逆元）

Invoker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 122768/62768 K (Java/Others)
Total Submission(s): 907    Accepted Submission(s): 364

Problem Description

On of Vance's favourite hero is Invoker, Kael. As many people knows Kael can control the elements and combine them to invoke a powerful skill. Vance like Kael very much so he changes the map to make Kael more powerful. 

In his new map, Kael can control n kind of elements and he can put m elements equal-spacedly on a magic ring and combine them to invoke a new skill. But if a arrangement can change into another by rotate the magic ring or reverse the ring along the axis, they will invoke the same skill. Now give you n and m how many different skill can Kael invoke? As the number maybe too large, just output the answer mod 1000000007.

 

 

Input

The first line contains a single positive integer T( T <= 500 ), indicates the number of test cases.
For each test case: give you two positive integers n and m. ( 1 <= n, m <= 10000 )

 

 

Output

For each test case: output the case number as shown and then output the answer mod 1000000007 in a line. Look sample for more information.

 

 

Sample Input

2 3 4 1 2

 

 

Sample Output

Case #1: 21 Case #2: 1

Hint

For Case #1: we assume a,b,c are the 3 kinds of elements. Here are the 21 different arrangements to invoke the skills / aaaa / aaab / aaac / aabb / aabc / aacc / abab / / abac / abbb / abbc / abcb / abcc / acac / acbc / / accc / bbbb / bbbc / bbcc / bcbc / bccc / cccc /

 

 

Source

2011 Multi-University Training Contest 9 - Host by BJTU

 

 

Recommend

xubiao

 

 

 

 

 

这题就是用polya定理，由于要取模，而且要除于一个数，所有要逆元素。

 

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <iostream>

#include <stdlib.h>

#include <time.h>

#include <math.h>

using namespace std;

const int MOD= 1e9+7;



long long pow_m(long long a,int n)

{

    long long ret = 1;

    long long temp = a%MOD;

    while(n)

    {

        if(n&1)

        {

            ret *= temp;

            ret %= MOD;

        }

        temp *= temp;

        temp %= MOD;

        n >>= 1;

    }

    return ret;

}

int gcd(int a,int b)

{

    if(b == 0)return a;

    return gcd(b,a%b);

}

//******************************

//返回d=gcd(a,b);和对应于等式ax+by=d中的x,y

long long extend_gcd(long long a,long long b,long long &x,long long &y)

{

    if(a==0&&b==0) return -1;//无最大公约数

    if(b==0){x=1;y=0;return a;}

    long long d=extend_gcd(b,a%b,y,x);

    y-=a/b*x;

    return d;

}

//*********求逆元素*******************

//ax = 1(mod n)

long long mod_reverse(long long a,long long n)

{

    long long x,y;

    long long d=extend_gcd(a,n,x,y);

    if(d==1) return (x%n+n)%n;

    else return -1;

}



int main()

{

    int T;

    int m,n;

    scanf("%d",&T);

    int iCase = 0;

    while(T--)

    {

        iCase++;

        scanf("%d%d",&m,&n);

        long long ans = 0;

        if(n%2==0)

        {

            ans = n/2*pow_m(m,n/2)+n/2*pow_m(m,n/2+1);

            ans %= MOD;

        }

        else ans = n*pow_m(m,n/2+1);

        //cout<<ans<<endl;

        for(int i = 0;i < n;i++)

        {

            ans += pow_m(m,gcd(i,n));

            ans %= MOD;

            //cout<<ans<<endl;

        }

        ans *= mod_reverse(2*n,MOD);

        ans%=MOD;

        printf("Case #%d: %I64d\n",iCase,ans);

    }

    return 0;

}