贪心水题

A - FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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 37 24 35 220 325 1824 1515 10-1 -1 

 

Sample Output

 13.33331.500 

 平均值排序；背包问题的变形

#include<stdio.h>#include<algorithm>using namespace std;struct node{double a;double b;double c;}s[1000];bool cmp(node a,node b){return a.c>b.c;}int main(){    double p,q;   int i,m,n;  while(scanf("%d%d",&m,&n),(m!=-1)&&(n!=-1)){ p=0; q=0; for(i=0;i<n;i++){scanf("%lf%lf",&s[i].a,&s[i].b);s[i].c=s[i].a/s[i].b;}sort(s,s+n,cmp);i=0;while(p<m&&i<n){p+=s[i].b;q+=s[i].a;i++;}printf("%.3f\n",q-(p-m)*(s[i-1].a/s[i-1].b));}}