HDU FatMouse' Trade
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.
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
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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 else 40 { 41 answer=answer+WH[j].JB; 42 m-=WH[j].CF; 43 } 44 } 45 printf("%.3f\n",answer); 46 } 47 return 0; 48 }