The warehouse has N rooms. The i-th room contains Jii pounds of JavaBeans and requires Fii pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get Jii* a% pounds of JavaBeans if he pays Fii* 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 Jii and Fii 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
31.500FatMouse 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 Jii pounds of JavaBeans and requires Fii pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get Jii* a% pounds of JavaBeans if he pays Fii* 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.
OutputFor 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 -1Sample Output
13.333 31.500
《贪心算法的应用，按单位javabean排序》
#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
struct Room
{
double j; //房间里含的总javabean
double f; //所需猫食
double ave; //单位交换的javabean
} room[10000];
int cmp(const Room a,const Room b)
{
return a.ave>b.ave;
}
int main()
{
int n,m;
while(cin >> m>>n&& (n!=-1||m!=-1))
{
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&room[i].j,&room[i].f);
room[i].ave=room[i].j/room[i].f;
}
sort(room,room+n,cmp); //单位价值排序
double sum=0;
int i=0;
while(m>0&&i<n)
{
if(room[i].f<m)
{
sum+=room[i].j;
m-=room[i].f; //全部能换走
}
else // 一定要注意猫粮很多，而猫的仓库很少的情况，不仅仅要判断猫粮完了没，还要判断仓库完了没。
{
sum+=m*room[i].ave; //只能部分换走
m=0;
}
i++;
}
printf("%.3lf\n",sum);
}
return 0;
}