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HDU FatMouse' Trade

2024-07-20 15:58:13   218人阅读

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20968    Accepted Submission(s): 6501


Problem Description
FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.
 

 

Input
The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1‘s. All integers are not greater than 1000.
 

 

Output
For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.
 

 

Sample Input
5 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
 

 

Sample Output
13.333 31.500
 

 

Author
CHEN, Yue
 

 

Source
ZJCPC2004
 

 

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JGShining
吐槽一下，竟然不支持lambda表达式...
 1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 using namespace std; 5 struct WareHouse 6 {  7     int JB,CF; 8     double ratio; 9 };10 bool cmp(WareHouse a,WareHouse b)11 {12     return a.ratio  >  b.ratio;13 }  14 int main()15 {16     int m,n;17     while(scanf_s("%d%d",&m,&n)&&m!=-1&&n!=-1)18     {19         double answer=0;20         //int*p = new int[n];21         vector<WareHouse> WH;22         int i=n;23         while(i--)24         {25             WareHouse temp;26             scanf_s("%d%d",&temp.JB,&temp.CF);27             temp.ratio = (double)temp.JB/temp.CF;28             WH.push_back(temp);29         }30     //    sort(WH.begin(),WH.end(),[](const WareHouse &a,const WareHouse&b ){ return a.ratio>b.ratio;});31         sort(WH.begin(),WH.end(),cmp);32         for (int j = 0; j < n; j++)33         {34             if(m<=WH[j].CF)35             {36                 answer=answer+m*WH[j].ratio;37                 m=0;38             }39             else40             {41                 answer=answer+WH[j].JB;42                 m-=WH[j].CF;43             }44         }45         printf("%.3f\n",answer);46     }47     return 0;48 }

 

HDU FatMouse' Trade


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