湖南大学ACM程序设计新生杯大赛（同步赛）H - Yuanyuan Long and His Ballons

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

题解

$dp$。

$dp[i][j]$表示在第$1$个人涂第一种颜色，涂完$i$个人，且第$i$个人涂第$j$种颜色的方案数。

$sum = dp[n][2]+...+dp[n][k]$，答案就是$sum*k$。

有很多优化可以搞，什么优化都没做就过了......

#include<cstdio>
using namespace std;

long long mod = 100000007LL;
long long dp[10010][110];

int main() {
  int T, n, k;
  scanf("%d", &T);
  while(T --) {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i ++) {
      for(int j = 1; j <= k; j ++) {
        dp[i][j] = 0;
      }
    }
    dp[1][1] = 1;
    for(int i = 2; i <= n; i ++) {
      long long sum = 0;
      for(int j = 1; j <= k; j ++) {
        sum = (sum + dp[i - 1][j]) % mod;
      }
      for(int j = 1; j <= k; j ++) {
        dp[i][j] = (sum - dp[i - 1][j] + mod) % mod;
      }
    }
    long long sum = 0;
    for(int j = 2; j <= k; j ++) {
      sum = (sum + dp[n][j]) % mod;
    }
    sum = sum * k % mod;
    printf("%lld\n", sum);
  }
  return 0;
}