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 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.
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 2 31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
#include<iostream>
using namespace std;
#include<math.h>
#include<string.h>
#define maxn 100000
long long d[maxn+];
int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
int main()
{
int t,n,m,i,j;
int V[],N[];
cin>>t;
while(t--)
{
memset(d,,sizeof(d));
cin>>n>>m;
for(i=;i<n;i++)
cin>>V[i];
for(j=;j<n;j++)
cin>>N[j];
for(i=;i<n;i++)
for(j=m;j>=N[i];j--)
{
d[j]=max(d[j],d[j-N[i]]+V[i]);
}
cout<<d[m]<<endl;
}
return ;
}