Zombie’s Treasure Chest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 994    Accepted Submission(s): 211


Problem Description
  Some brave warriors come to a lost village. They are very lucky and find a lot of treasures and a big treasure chest, but with angry zombies.
  The warriors are so brave that they decide to defeat the zombies and then bring all the treasures back. A brutal long-drawn-out battle lasts from morning to night and the warriors find the zombies are undead and invincible.
  Of course, the treasures should not be left here. Unfortunately, the warriors cannot carry all the treasures by the treasure chest due to the limitation of the capacity of the chest. Indeed, there are only two types of treasures: emerald and sapphire. All of the emeralds are equal in size and value, and with infinite quantities. So are sapphires.
  Being the priest of the warriors with the magic artifact: computer, and given the size of the chest, the value and size of each types of gem, you should compute the maximum value of treasures our warriors could bring back.
 

 

Input
  There are multiple test cases. The number of test cases T (T <= 200) is given in the first line of the input file. For each test case, there is only one line containing five integers N, S1, V1, S2, V2, denoting the size of the treasure chest is N and the size and value of an emerald is S1 and V1, size and value of a sapphire is S2, V2. All integers are positive and fit in 32-bit signed integers.
 

 

Output
  For each test case, output a single line containing the case number and the maximum total value of all items that the warriors can carry with the chest.
 

 

Sample Input
2 100 1 1 2 2 100 34 34 5 3
 

 

Sample Output
Case #1: 100 Case #2: 86
 

 

Source
2011 Asia Shanghai Regional Contest
 

 

Recommend
lcy
 
 
 
#include<stdio.h>
#include<algorithm>
using namespace std;
long long gcd(long long da,long long xiao)
{
    long long temp;
    while(xiao!=0)
    {
        temp=da%xiao;
        da=xiao;
        xiao=temp;
    }    
    return da;
}    
int main()
{
    int iCase=0;
    int T;
    scanf("%d",&T);
    long long N,S1,V1,S2,V2;
    while(T--)
    {
        iCase++;
        scanf("%I64d%I64d%I64d%I64d%I64d",&N,&S1,&V1,&S2,&V2);
        long long tmp=S1*S2/gcd(S1,S2);
        long long res;
        long long tt=N/tmp;
        N=N%tmp;
        if(tt)
        {
            tt--;
            N+=tmp;
        }    
        res=max((tt)*(tmp/S1)*V1,(tt)*(tmp/S2)*V2);
        long long res2=0;
        if(S2>S1)
        {
            long long t;
            t=S1;S1=S2;S2=t;
            t=V1;V1=V2;V2=t;
        }    
        for(int i=0;i<=N/S1;i++)
        {
            if(res2<i*V1+(N-i*S1)/S2*V2)res2=i*V1+(N-i*S1)/S2*V2;
        }    
        res+=res2;
        printf("Case #%d: %I64d\n",iCase,res);
    }    
    return 0;
}

文章来源:http://www.cnblogs.com/kuangbin/archive/2011/11/07/2240182.html