湖南大学ACM程序设计新生杯大赛 - Yuanyuan Long and His Ballons（环形染色问题）

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述 

Yuanyuan Long is a dragon like this picture?


                                   


I don’t know, maybe you can ask him. But I’m sure Yuanyuan Long likes ballons, he has a lot of ballons.           

One day, he gets n white ballons and k kinds of pigment, and he thought a problem: 

1.      Number these ballons b1, b2,  … , bi, …,  to bn.

2.      Link these ballons to a circle in order, and color these ballons.

3.      He need to make sure that any neighbor ballons have different colors. 

He wants to know how many solutions to solve this problem. Can you get the answer to him? The answer maybe very large, get the answer MOD 100000007.

For Example: with 3 ballons and 3 kinds of pigment



Your answer is 3*2*1 MOD 100000007 = 6. 
The answer maybe large, you can use integer type long long. 

输入描述:

The first line is the cases T. ( T <=
100)
For next T lines, each line contains n and
k. (2<= n <= 10000,  2<= k
<=100)

输出描述:

For each test case, output the answer on
each line.

示例1

输入

3
3 3
4 2
5 3

输出

6
2
30

#include <set>
#include <map>
#include <math.h>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
const double pi = acos(-1);
const LL maxn = 1000100;
const LL mod = 100000007;
//排列组合，环形染色问题
//
//一般结论为
//如果有M个颜色，n个区域，可以有如下公式（可以通过数列求和推导）
//
//（m-1）^n+(m-1)(-1)^n

LL qkm(LL base,LL mi)
{
LL ans=1;
while(mi){
if(mi&1) ans=ans*base;
ans%=mod;
base=base*base;
base%=mod;
mi>>=1;
}
return ans;
}

int main()
{
LL T;
scanf("%lld",&T);
while (T--) {
LL n,m;
scanf("%lld %lld",&n,&m);
printf("%lld\n",qkm((m-1),n)+(m-1)*qkm(-1,n));
}

}