DP

　　这种题居然是DP……原来统计方案数是可以用动态规划来做的啊= =用一些变量来维护一些信息，保证方案可行性/合法性

　　人太弱实在是有些忧伤……

 1 /**************************************************************
 2     Problem: 1037
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:460 ms
 7     Memory:42660 kb
 8 ****************************************************************/
 9
10 //BZOJ 1037
11 #include<vector>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<iostream>
16 #include<algorithm>
17 #define rep(i,n) for(int i=0;i<n;++i)
18 #define F(i,j,n) for(int i=j;i<=n;++i)
19 #define D(i,j,n) for(int i=j;i>=n;--i)
20 #define pb push_back
21 using namespace std;
22 inline int getint(){
23     int v=0,sign=1; char ch=getchar();
24     while(ch<‘0‘||ch>‘9‘){ if (ch==‘-‘) sign=-1; ch=getchar();}
25     while(ch>=‘0‘&&ch<=‘9‘){ v=v*10+ch-‘0‘; ch=getchar();}
26     return v*sign;
27 }
28 const int N=155,INF=~0u>>2,MOD=12345678;
29 typedef long long LL;
30 /******************tamplate*********************/
31 int n,m,k,f[N][N][21][21],ans;
32 int main(){
33 #ifndef ONLINE_JUDGE
34     freopen("1037.in","r",stdin);
35     freopen("1037.out","w",stdout);
36 #endif
37     n=getint(); m=getint(); k=getint();
38     f[0][0][0][0]=1;
39     F(i,0,n) F(j,0,m) F(k1,0,k) F(k2,0,k){
40         if (k1+1<=k) (f[i+1][j][k1+1][max(k2-1,0)]+=f[i][j][k1][k2])%=MOD;
41         if (k2+1<=k) (f[i][j+1][max(k1-1,0)][k2+1]+=f[i][j][k1][k2])%=MOD;
42     }
43     F(x,0,k) F(y,0,k) (ans+=f[n][m][x][y])%=MOD;
44     printf("%d\n",ans);
45     return 0;
46 }