C. Colorful Bricks(dp)

最新推荐文章于 2022-04-01 19:15:18 发布

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最新推荐文章于 2022-04-01 19:15:18 发布

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分类专栏： codeforces 日常训练 DP

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本文介绍了一个关于计数的问题，Chouti在空闲时给砖块涂色，他有n块砖和m种颜色。涂色完成后，他发现有k块砖的颜色与其左侧不同。任务是计算有多少种不同的涂色方式，使得至少有一块砖的颜色与其他相邻砖不同。题目要求计算结果对998244353取模。

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On his free time, Chouti likes doing some housework. He has got one new task, paint some bricks in the yard.

There are ?n bricks lined in a row on the ground. Chouti has got ?m paint buckets of different colors at hand, so he painted each brick in one of those ?m colors.

Having finished painting all bricks, Chouti was satisfied. He stood back and decided to find something fun with these bricks. After some counting, he found there are ?k bricks with a color different from the color of the brick on its left (the first brick is not counted, for sure).

So as usual, he needs your help in counting how many ways could he paint the bricks. Two ways of painting bricks are different if there is at least one brick painted in different colors in these two ways. Because the answer might be quite big, you only need to output the number of ways modulo 998244353998244353.

Input

The first and only line contains three integers ?n, ?m and ?k (1≤?,?≤2000,0≤?≤?−11≤n,m≤2000,0≤k≤n−1) — the number of bricks, the number of colors, and the number of bricks, such that its color differs from the color of brick to the left of it.

Output

Print one integer — the number of ways to color bricks modulo 998244353998244353.

思路：设dp[i][j]表示第i位有j个不同的方法数。 dp[i][j]=dp[i-1][j]+dp[i-1][j-1]*(m-1);（dp[i][j]=上一状态颜色不同为j-1的方法数+上一状态颜色不同为j-1的方法数乘以m-1）。

# include <iostream>
# include <cstdio>
# include <cmath>
#define ll long long
# define pi acos(-1.0)
#define mod 998244353
using namespace std;
ll dp[2008][2008];
int main(){
    ll n,m,k;
    cin>>n>>m>>k;
    for(int i=1;i<=k;i++){
        dp[1][i]=0;
    }
    for(int i=1;i<=n;i++)
        dp[i][0]=m;
    for(int i=2;i<=n;i++){
        for(int j=1;j<=k;j++){
            dp[i][j]=(dp[i-1][j]+(dp[i-1][j-1]*(m-1))%mod)%mod;
        }
    }
    cout<<dp[n][k]<<endl;
    return 0;
}