HDU FatMouse' Trade
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FatMouse' Trade
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
Author
CHEN, Yue
Source
ZJCPC2004
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn=1005;
int dp[maxn];
struct aa{
float a,b,v;}a[maxn];
int cmp(struct aa a,struct aa b){
return a.v>b.v;}
int main(){//freopen("f.txt","r",stdin);
int m,n;
while(scanf("%d%d",&m,&n)){
if(m==-1||n==-1) break;
int i;
for(i=0;i<n;i++)
scanf("%f%f",&a[i].a,&a[i].b),a[i].v=a[i].a/a[i].b;
sort(a,a+n,cmp);
float s=0;
m=(float )m;
for(i=0;i<n;i++){
if(m-a[i].b>=0){
s+=a[i].a;
m-=a[i].b;
}
else{
s+=m*a[i].v;
break;
}
}
printf("%.3f\n",s);
}
}
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