hdoj1009FatMouse'트레이드(욕심 풀이)

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
#include<stdio.h> #include<stdlib.h> double J[1005],F[1005]; struct node { double s; int k; }S[1000]; int cmp(const void* p1,const void* p2) { return ((node *)p2)->s > ((node *)p1)->s ? 1 : -1;// , ((node*)p2)->s-((node*)p1)->s; } int main() { double m; int n; while(scanf("%lf%d",&m,&n)!=EOF) { if(m==-1&&n==-1) break; for(int i=0;i<n;i++) { scanf("%lf%lf",&J[i],&F[i]); S[i].s=J[i]/F[i]; S[i].k=i; } qsort(S,n,sizeof(S[0]),cmp); double f=0; int j=0; for(int j=0;j<n;j++) { if(F[S[j].k]<=m) { f=f+J[S[j].k]; m=m-F[S[j].k]; } else { double k1=m/F[S[j].k]; f=f+J[S[j].k]*k1; break; } } printf("%.3lf/n",f); } return 0; }

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