题解 | #樱花#

#include <bits/stdc++.h>

#define ll long long 
#define int ll

using namespace std;

const int N = 1010, mod = 10007;
int dp[2][N][N], k;
int a, b, n, m;

signed main()
{
	cin >> a >> b >> k >> n >> m;
	k -- ;
	
	
	dp[0][1][0] = a, dp[0][0][1] = b;
	for(int mark = 1; mark <= k; mark ++ ) {
		for(int i = 0; i <= min(mark + 1, m); i ++ ) {
			for(int j = 0; j <= min(mark + 1, m); j ++ ) {
				if(i - 1 >= 0) dp[mark & 1][i][j] = (dp[mark & 1][i][j] + dp[(mark - 1) & 1][i - 1][j] * a % mod) % mod;
				if(j - 1 >= 0) dp[mark & 1][i][j] = (dp[mark & 1][i][j] + dp[(mark - 1) & 1][i][j - 1] * b % mod) % mod;
			}
		}
	}
	printf("%lld\n", dp[k & 1][n][m] % mod);
	return 0;
}