太坑了这个题,还要考虑不花钱就可以兑换,哇塞,便宜都让老鼠给占了
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
代码:
#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
double a,b,v;
}s[100010];
bool cmp(struct node a,struct node b)
{
return a.v>b.v;
}
int main()
{
int n,i,j;
double m,k;
while(scanf("%lf%d",&m,&n)!=EOF&&(m!=-1||n!=-1))
{
k=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&s[i].a,&s[i].b);
if(s[i].b==0)
s[i].v=s[i].a;
else
s[i].v=s[i].a/s[i].b;
}
sort(s,s+n,cmp);
for(i=0;i<n;i++)
{
if(s[i].b<=m)
{
m-=s[i].b;
k+=s[i].a;
}
else
{
if(m!=0)
k+=s[i].v*m;
m=0;
}
}
printf("%.3lf\n",k);
}
return 0;
}