HDU4091(2011 Asia Shanghai Regional Contest)

bigbigship發表於2014-07-23
AI

Zombie’s Treasure Chest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4085    Accepted Submission(s): 836


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
線性規劃題目，由於只能取整點 最優點可能取不到 ，因此索性將容量分為兩部分：
1）可以恰好裝翡翠或者藍寶石的
2）剩下的部分
第一部分直接取兩者總利潤的最大值
第二部分直接列舉 看取那種最好
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int cas;
    __int64 n,s1,v1,s2,v2;
    scanf("%d",&cas);
    for(int j=1;j<=cas;j++){
        scanf("%I64d%I64d%I64d%I64d%I64d",&n,&s1,&v1,&s2,&v2);
        __int64 tot=n/(s1*s2);
        __int64  n1=tot*(s1*s2);
        __int64 val=0;
        if(tot*(s1*s2)/s1*v1>tot*(s1*s2)/s2*v2)
            val=tot*(s1*s2)/s1*v1;
        else
            val=tot*(s1*s2)/s2*v2;
        __int64 val2=0;
        __int64  n2=n-n1;
        if(s1>s2){
          for(int i=0;i*s1<=n2;i++){
            __int64 ans=i*v1+(n2-i*s1)/s2*v2;
            if(val2<ans)
                val2=ans;
          }
        }
        else{
          for(int i=0;i*s2<=n2;i++){
            __int64 ans=i*v2+(n2-i*s2)/s1*v1;
            if(ans>val2)
                val2=ans;
          }
        }
        //cout<<n1<<" "<<n2<<endl;
        //cout<<val<<" "<<val2<<endl;
        printf("Case #%d: %I64d\n",j,val+val2);
    }
    return 0;
}










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