[ZJOI2008]生日聚会

题解:

我们设dp[a][b][c][d]表示目前已经有a个男生,b个女生,在某一时刻男生最多比女生多c个,在某一时刻女生最多比男生多d个.
所以就可以转移了:
(dp[a+1][b][c+1][max(d-1,0)]+=dp[a][b][c][d])%=mod;
(dp[a][b+1][max(c-1,0)][d+1]+=dp[a][b][c][d])%=mod;

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#define MAXN 500010
#define RG register
#define LL long long int
using namespace std;
const int INF=1e9;
const int mod=12345678;
int n,m,k;
LL dp[155][155][25][25];
LL ans;
int main()
{
  freopen("1.in","r",stdin);
  scanf("%d%d%d",&n,&m,&k);
  dp[0][0][0][0]=1;
  for(int a=0;a<=n;a++)
    for(int b=0;b<=m;b++)
      for(int c=0;c<=k;c++)
    for(int d=0;d<=k;d++)
      {
        (dp[a+1][b][c+1][max(d-1,0)]+=dp[a][b][c][d])%=mod;
        (dp[a][b+1][max(c-1,0)][d+1]+=dp[a][b][c][d])%=mod;
      }
  for(int c=0;c<=k;c++)
    for(int d=0;d<=k;d++)
      (ans+=dp[n][m][c][d])%=mod;
  printf("%lld
",ans);
  return 0;
}