bzoj 2281: [Sdoi2011]黑白棋

再次，，，，，虚（一开始看错题了，看成一次移动一个棋子，能移动1-d个格子。。。这样的话有没有大神会做？？本蒟蒻就教）

额，，直接%%%%把。。。http://hzwer.com/5760.html

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#include<queue>
#define N 1000005
#define inf 1000000000
#define LL long long
using namespace std;
const LL mod=1e9+7;
LL tot,ans;
int n,K,d,p;
LL bin[25];
LL c[10005][205],f[25][10005];
void pre()
{
    for (int i=0; i<=n; i++) c[i][0]=1;
    for (int i=1; i<=n; i++)
        for (int j=1; j<=min(2*K,i); j++)
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
}
int C(int x, int y)
{
    if (y>x-y) y=x-y;
    return c[x][y];
}
void add(LL &x, LL y)
{
    x=(x+y)%mod;
}
int main()
{
    bin[0]=1; for (int i=1; i<=15; i++) bin[i]=bin[i-1]<<1;
    scanf("%d %d %d",&n,&K,&d); K/=2; pre(); f[0][0]=1;
    for (int i=0; i<15; i++)
        for (int j=0; j<=n-2*K; j++)
            for (int k=0; k*(d+1)<=K && j+(d+1)*k*bin[i]<=n-2*K; k++)
            {
                add(f[i+1][j+k*(d+1)*bin[i]],f[i][j]*C(K,k*(d+1)));
            }
    for (int i=0; i<=n-2*K; i++)
        add(ans,f[15][i]*C(n-i-K,K));
    tot=C(n,K*2);
    cout<<(tot+mod-ans)%mod;
    return 0;
}