[ZJOI2008]生日聚会

dp，f[i][j][k][l]表示前i个人，j个男生，后缀中男生比女生最多多k人，最少少l人的方案数。

/**
 * Problem:Party
 * Author:Shun Yao
 * Time:2013.5.30
 * Result:Accepted
 * Memo:DP
 */

#include <cstring>
#include <cstdlib>
#include <cstdio>

long min(long x, long y) {
	return x < y ? x : y;
}
long max(long x, long y) {
	return x > y ? x : y;
}

const long MOD = 12345678;

long n, m, lim, N, f[20043891], ans;

int main() {
	freopen("party.in", "r", stdin);
	freopen("party.out", "w", stdout);
	scanf("%ld%ld%ld", &n, &m, &lim);
	N = n + m;
	memset(f, 0, sizeof f);
#define F(a, b, c, d) f[(((a) * (n + 1) +(b)) * (lim + 1) + (c)) * (lim + 1) + (d)]
	long i, j, k, l, J, K, L;
	F(0, 0, 0, 0) = 1;
	for (i = 0; i < N; ++i) {
		for (J = min(i, n), j = 0; j <= J; ++j)
			for (K = min(j, lim), k = 0; k <= K; ++k)
				for (L = min(i - j, lim), l = 0; l <= L; ++l) {
					if (k < lim && j < n) {
						long &a = F(i + 1, j + 1, k + 1, max(l - 1, 0));
						a = (a + F(i, j, k, l)) % MOD;
					}
					if (l < lim && i - j < m) {
						long &b = F(i + 1, j, max(k - 1, 0), l + 1);
						b = (b + F(i, j, k, l)) % MOD;
					}
				}
	}
	ans = 0;
	for (k = 0; k <= lim; ++k)
		for (l = 0; l <= lim; ++l)
			ans = (ans + F(N, n, k, l)) % MOD;
	printf("%ld", ans);
	fclose(stdin);fclose(stdout);
	return 0;
}