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;
}