The first and only line contains three integers nn, mm and kk (1≤n,m≤2000,0≤k≤n−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.

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

3 3 0

3

3 2 1

4

In the first example, since k=0k=0, the color of every brick should be the same, so there will be exactly m=3m=3 ways to color the bricks.

In the second example, suppose the two colors in the buckets are yellow and lime, the following image shows all 44 possible colorings.

題意概括：

N 個 方塊， M 種顏色，存在 K 個方塊使得它與相鄰左邊的方塊顏色不同。

求塗色方案數。

解題思路：

換角度思考，其實就是 求把 N塊方塊分成 K+1塊與相鄰左邊塗色不同的方案數。

楊輝三角求組合數 C(N-1, K), 因為第一塊不考慮與左邊顏色的關係 有 M 種可能，其餘的都要去掉左邊那一塊的顏色，所以只有 M-1種可能，即 M*（M-1）*（M-1）*......*(M-1) ;

分塊方案數 * 顏色方案數 即最後答案。

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 #include <cmath>
 7 #define INF 0x3f3f3f3f
 8 #define LL long long
 9 using namespace std;
10 const int MAXN = 2e3+10;
11 const LL MOD = 998244353;
12 LL c[MAXN][MAXN];
13 
14 LL qpow(LL x, LL n)
15 {
16     LL res = 1LL;
17     while(n){
18         if(n&1) res = ((res%MOD*x%MOD)+MOD)%MOD;
19         x = x*x%MOD;
20         n>>=1LL;
21     }
22     return res;
23 }
24 
25 int main()
26 {
27     LL N, M, K;
28     cin >> N >> M >> K;
29     //memset(c, 1LL, sizeof(1LL));
30     c[0][0] = c[1][0] = c[1][1] = 1LL;
31 
32     for(int i = 2; i <= N; i++){
33         c[i][0] = 1LL;
34         for(int j = 1; j < i; j++){
35             c[i][j] = (c[i-1][j-1] + c[i-1][j])%MOD;
36         }
37         c[i][i] = 1LL;
38     }
39 
40     //cout << c[N-1][K];
41 
42     LL ans = 1LL;
43     ans = (M%MOD*c[N-1][K]%MOD*qpow(M-1LL, K)%MOD + MOD)%MOD;
44     cout << ans << endl;
45     return 0;
46 
47 }