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 ?

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.

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

Sample Output

 #include<bits/stdc++.h>

 using namespace std;

 const int maxn=;

 int t,n,W,v[maxn],w[maxn],dp[maxn][maxn];

 int main(){

     while(cin>>t){

         while(t--){

             cin>>n>>W;

             for(int i=;i<=n;++i)cin>>v[i];

             for(int i=;i<=n;++i)cin>>w[i];

             memset(dp,,sizeof(dp));

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

                 for(int j=;j<=W;++j){

                     if(j<w[i])dp[i][j]=dp[i-][j];//无法挑选这个物品

                     else dp[i][j]=max(dp[i-][j],dp[i-][j-w[i]]+v[i]);//拿和不拿的两种情况都试一下

                 }

             }

             cout<<dp[n][W]<<endl;

         }

     }

     return ;

 }

 #include<bits/stdc++.h>

 using namespace std;

 int value[],weight[],dp[];//dp数组始终记录当前体积的最大价值

 int main()

 {

     int T,N,V;

     cin>>T;

     while(T--){

         cin>>N>>V;

         for(int i=;i<N;i++)cin>>value[i];//输入价值

         for(int i=;i<N;i++)cin>>weight[i];//输入体积

         memset(dp,,sizeof(dp));//初始化

         for(int i=;i<N;i++){    //个数

             for(int j=V;j>=weight[i];j--) //01背包

                 dp[j]=max(dp[j],dp[j-weight[i]]+value[i]); //比较放入i物体后的价值与不放之前的价值，记录大的值

         }

         cout<<dp[V]<<endl;//输出总体积的最大价值

     }

     return ;

 }