Pick apples

Time Limit: 1000MS Memory Limit: 165536KB

   

Problem Description

 

Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she has never seen so many apples before. Each kind of apple has a size and a price to be sold. Now the little girl wants to gain more profits, but she does not know how. So she asks you for help, and tell she the most profits she can gain.

 

Input

In the first line there is an integer T (T <= 50), indicates the number of test cases.

In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S <= 100) and the price (1 <= P <= 10000) of this kind of apple.

In the fourth line there is an integer V，(1 <= V <= 100,000,000)indicates the volume of the girl's bag.

Output

For each case, first output the case number then follow the most profits she can gain.

Example Input

11 12 13 16

Example Output

Case 1: 6

Hint

 

Author

2012年"浪潮杯"山东省第三届ACM大学生程序设计竞赛

/** @Author: lyuc* @Date:   2017-04-27 16:06:19* @Last Modified by:   lyuc* @Last Modified time: 2017-04-27 19:27:24*/#include 
      
       #include 
       
        #include 
        
         #include 
         
          #define LL long long#define maxn 1000000using namespace std;struct node{    int s,p;    bool operator <(const node & other) const {        return p*1.0/s>other.p*1.0/other.s;    }}fr[4];LL dp[1000005];int t;int v;LL res;int main(){    // freopen("in.txt","r",stdin);    scanf("%d",&t);    for(int ca=1;ca<=t;ca++){        printf("Case %d: ",ca);        res=0;        memset(dp,0,sizeof dp);        for(int i=0;i<3;i++){            scanf("%d%d",&fr[i].s,&fr[i].p);        }        scanf("%d",&v);        if(v<=maxn){            for(int i=0;i<3;i++){                for(int j=fr[i].s;j<=v;j++){                    dp[j]=max(dp[j],dp[j-fr[i].s]+fr[i].p);                }            }            printf("%lld\n",dp[v]);        }else{            sort(fr,fr+3);            while(v>maxn){                res+=fr[0].p;                v-=fr[0].s;            }            for(int i=0;i<3;i++){                for(int j=fr[i].s;j<=v;j++){                    dp[j]=max(dp[j],dp[j-fr[i].s]+fr[i].p);                }            }            printf("%lld\n",res+dp[v]);        }    }        return 0;}