c++ style memset its for get strong IE HA

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 1 5 10 1 2 3 4 5 5 4 3 2 1 Sample Output 14 題意 最裸的01背包，給你背包總量和物品數，以及物品的價值和體積，讓你求背包裝滿後的最大價值

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,N,V,i,j,w[1005],v[1005
 
],dp[1005];
    cin>>t;
    while(t--)
    {
        memset(dp,0,sizeof dp);
        cin>>N>>V;
        for(i=1;i<=N;i++)
            scanf("%d",&v[i]);
        for(i=1;i<=N;i++)
            scanf("%d",&w[i]);
        for(i=1;i<=N;i++)        //每一個骨頭
            for(j=V;j>=w[i];j--) //每個骨頭的價值 
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]); 
        
        cout<<dp[V]<<endl;
    }
    return 0;
}

【HDU 2602】Bone Collector（裸的01背包）