HDU 3923 Invoker

Invoker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 122768/62768 K (Java/Others)
Total Submission(s): 2013    Accepted Submission(s): 928

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 /

 

题意

用m种颜色给n个珠子染色，两个方案不同当且仅当满足旋转或者翻转后不同，求一共有多少不同的方案。

 

分析

polya计数原理，和poj1286差不多。

答案对1e9+7取模，最后在除以总的置换数时，需要用到逆元。

code

#include<cstdio>

typedef long long LL;
const long long mod = 1e9+7; //-

LL gcd(LL a,LL b) {
    if (b==0) return a;
    return gcd(b,a%b);
}
LL ksm(LL a,LL b) {
    LL ans = 1;
    while (b) {
        if (b & 1) ans = (ans * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans;
}
LL Polya(LL n,LL m) {
    LL ans = 0;
    for (LL i=1; i<=n; ++i) 
        ans = (ans + ksm(m,gcd(n,i))) % mod;
    if (n & 1) ans = (ans + n * ksm(m,(n-1)/2+1)) % mod;
    else {
        ans = (ans + n/2*ksm(m,(n-2)/2+2)) % mod;
        ans = (ans + n/2*ksm(m,n/2)) % mod;    
    }
    ans = (ans * ksm(n*2,mod-2)) % mod;
    return ans;
}
int main () {
    int t;LL n,m;
    scanf("%d",&t);
    for (int c=1; c<=t; ++c) {
        scanf("%lld%lld",&m,&n); // -
        printf("Case #%d: %lld\n",c,Polya(n,m));
    }
    return 0;
}