[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;

 1 #include<iostream>
 2 #include<cstdlib>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<cstring>
 7 #include<queue>
 8 #include<vector>
 9 #include<set>
10 #define MAXN 500010
11 #define RG register
12 #define LL long long int
13 using namespace std;
14 const int INF=1e9;
15 const int mod=12345678;
16 int n,m,k;
17 LL dp[155][155][25][25];
18 LL ans;
19 int main()
20 {
21   freopen("1.in","r",stdin);
22   scanf("%d%d%d",&n,&m,&k);
23   dp[0][0][0][0]=1;
24   for(int a=0;a<=n;a++)
25     for(int b=0;b<=m;b++)
26       for(int c=0;c<=k;c++)
27     for(int d=0;d<=k;d++)
28       {
29         (dp[a+1][b][c+1][max(d-1,0)]+=dp[a][b][c][d])%=mod;
30         (dp[a][b+1][max(c-1,0)][d+1]+=dp[a][b][c][d])%=mod;
31       }
32   for(int c=0;c<=k;c++)
33     for(int d=0;d<=k;d++)
34       (ans+=dp[n][m][c][d])%=mod;
35   printf("%lld\n",ans);
36   return 0;
37 }

 

posted @ 2017-12-21 21:48  楼主大大  阅读(125)  评论(0)  编辑  收藏  举报