BZOJ1037

传送门：BZOJ1037

令

当然要在合法的前提下进行。

答案是

答案忘记模了调了半天囧…

代码上的小细节见下。

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int p=12345678;
int ans;
int f[305][155][25][25];
int n,m,c;

void Solve()
{
    f[0][0][0][0]=1;
    for(int i=0;i<n+m;i++)
        for(int j=0;j<=n;j++)
            for(int k=0;k<=c;k++)
                for(int l=0;l<=c;l++)
                    if(f[i][j][k][l])
                    {
                        if((i-j+1<=m)&&(l+1<=c)){
                            f[i+1][j][max(k-1,0)][l+1]+=f[i][j][k][l];
                            f[i+1][j][max(k-1,0)][l+1]%=p;
                        }
                        if((j+1<=n)&&(k+1<=c)){
                            f[i+1][j+1][k+1][max(l-1,0)]+=f[i][j][k][l];
                            f[i+1][j+1][k+1][max(l-1,0)]%=p;
                        }
                    }
    for(int i=0;i<=c;i++)
        for(int j=0;j<=c;j++){
            ans+=f[n+m][n][i][j];
            ans%=p;
        }
    printf("%d\n",ans);
}

void Readdata()
{
    freopen("loli.in","r",stdin);
    scanf("%d%d%d",&n,&m,&c);
}

void Close()
{
    fclose(stdin);
    fclose(stdout);
}

int main()
{
    Readdata();
    Solve();
    Close();
    return 0;
}