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FatMouse’ Trade
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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Problem Description
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
Output
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
//本题是简单贪心问题。意思是老鼠有m榜猫粮，通过给猫粮东西data[i].b能够兑换data[i].a，求最多能够兑换多少食物。//仅仅需将data[i].a/data[i].b的值从大到小排序就可以求解。
#include<stdio.h>struct st{ double a; double b; double c;}data[1000];int main(){ int i,j,m,n; struct st data[1000],t; while(scanf("%d %d",&m,&n)&&(m!=-1||n!=-1)) { double sum=0.000; for(i=0;i<n;i++) { scanf("%lf %lf",&data[i].a,&data[i].b); } for(i=0;i<n;i++) { for( j=i+1;j<n;j++) { if((data[i].a/data[i].b)<data[j].a/data[j].b) { t=data[i]; data[i]=data[j]; data[j]=t;} } } for(i=0;i<n;i++) { if(m-data[i].b>=0.001) { sum+=data[i].a; m-=data[i].b; } else { sum=sum+m*data[i].a/data[i].b; break; } } printf("%.3lf\n",sum); } return 0;}
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