湖南大学ACM 程序设计竞赛 同步赛

链接：https://www.nowcoder.net/acm/contest/55/H
来源：牛客网

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.

  思路:用m种颜色在n个区域涂色   公式 (m-1)^n+(m-1)(-1)^n

# include 
# include 
# include 
using namespace std;
# define min(x,y)  ((x)<(y)?(x):(y))
# define max(x,y)    ((x)>(y)?(x):(y))
const int maxn=1e6+112;
const int mod = 100000007 ;
typedef long long ll;
int n;
int a[1005],b[1005];
int vis[maxn];
// 用m种颜色在n个区域涂色   公式 (m-1)^n+(m-1)(-1)^n
int main()
{
	int n,T,k;
	cin>>T;
    while(T--){
       cin>>n>>k;
       int m = k - 1;
       ll ans = 1;
       for(int i=1;i<=n;i++){
           ans = ans*m%mod;
       }
       if(n&1){
         ans = ans - m;
       }else{
         ans = ans + m;
       }
       cout<