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+;
 LL tot,ans;
 int n,K,d,p;
 LL bin[];
 LL c[][],f[][];
 void pre()
 {
     for (int i=; i<=n; i++) c[i][]=;
     for (int i=; i<=n; i++)
         for (int j=; j<=min(*K,i); j++)
             c[i][j]=(c[i-][j]+c[i-][j-])%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[]=; for (int i=; i<=; i++) bin[i]=bin[i-]<<;
     scanf("%d %d %d",&n,&K,&d); K/=; pre(); f[][]=;
     for (int i=; i<; i++)
         for (int j=; j<=n-*K; j++)
             for (int k=; k*(d+)<=K && j+(d+)*k*bin[i]<=n-*K; k++)
             {
                 add(f[i+][j+k*(d+)*bin[i]],f[i][j]*C(K,k*(d+)));
             }
     for (int i=; i<=n-*K; i++)
         add(ans,f[][i]*C(n-i-K,K));
     tot=C(n,K*);
     cout<<(tot+mod-ans)%mod;
     return ;
 }