Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 231
).

Sample Input

15 101 2 3 4 55 4 3 2 1

Sample Output

14

PS

简单的DP入门题

//
//  main.cpp
//  DP或贪心
//
//  Created by jetviper on 2017/3/18.
//  Copyright © 2017年 jetviper. All rights reserved.
//
#include <iostream>
#include<cstdio>
using namespace std;

int main()
{
    int t,n,v,w[1024],p[1024];
    int dp[1024];
    scanf("%d",&t);
for(int q=0;q<t;q++){
memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&v);
    for(int i=0;i<n;i++)scanf("%d",&p[i]);//输入物品价值
    for(int i=0;i<n;i++)scanf("%d",&w[i]);//输入物品体积

    for(int i=0;i<n;i++){
        for(int j=v;j>=w[i];j--){//背包填不满，所以此处为倒序遍历
           if(dp[j-w[i]]+p[i]>dp[j]){
                dp[j]=dp[j-w[i]]+p[i];
           }
        }

    }
cout<<dp[v]<<endl;

}
    return 0;
}