FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63198 Accepted Submission(s): 21342
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
简单贪心题，按性价比排序后优先取大的
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include<vector>
#include <map>
using namespace std;
#define inf 0x3f3f3f3f
struct node
{
int a,b;
double c;
} p[10000];
bool cmp(node a,node b)
{
return a.c>b.c;
}
int main()
{
int m,n;
while(~scanf("%d%d",&m,&n))
{
if(m==-1&&n==-1)
break;
for(int i=0; i<n; i++)
{
scanf("%d%d",&p[i].a,&p[i].b);
p[i].c=p[i].a*1.0/p[i].b;
}
sort(p,p+n,cmp);
double ans=0;
for(int i=0;i<n;i++)
{
if(m>p[i].b)
{
ans+=p[i].a;
m-=p[i].b;
}
else
{
ans+=m*p[i].c;
break;
}
}
printf("%.3f\n",ans);
}
return 0;
}