HDU2602 DP + 裸 + 01背包

1）

#include <iostream>#include <string.h>using namespace std;const int maxn=1010;int value[maxn];int volume[maxn];long long int dp[maxn];int main(){    int kase;    cin>>kase;    while(kase--){        memset(dp,0,sizeof(dp));        int n,v;        cin>>n>>v;        for(int i=1;i<=n;i++){            cin>>value[i];        }        for(int i=1;i<=n;i++){            cin>>volume[i];        }            for(int i=1;i<=n;i++){            for(int j=v;j>=volume[i];j--){                dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);            }        }        cout<<dp[v]<<endl;    }}

2）

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