[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;

}