一道动态规划题

//dp[i][j]=(a,b) 表示前i个球中选j个球所需的最少盒子为a个加附加的b空间  b<=V

#include <iostream>

#include <cstring>

inline int Rint() {int x;scanf("%d%d",&x);return x;}

const int N=501,one=(1<<10),low=one-1,high=~low;

int dp[N][N],n,m,V,v[N],i,j,t,ans;

#define  min(x,y) (x<y?x:y)



inline int addl(int a, int b) {

	if ((a&low) + b <= V)

		return a + b;

	else return (a&high)+one + b;

}



int main() {

	for (int cas=Rint();cas--;) {

		n=Rint(),m=Rint(),V=Rint();

		m=(m<<10);//(m,0)

		for (i=1;i<=n;++i) v[i]=Rint();

		for (i=1;i<=n;++i) {

			dp[i][0]=0; //(0,0)

			for (j=1;j<i;++j) {

				dp[i][j]=min(dp[i-1][j],addl(dp[i-1][j-1],v[i]));

			}

			dp[i][i]=addl(dp[i-1][j-1],v[i]);

		}

		for (ans=0,i=1;i<=n;++i) if (dp[n][i]<m&&i>ans) ans=i;

		printf("%d\n",ans);

	}

	return 0;

}