HDU 1009：FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 75348    Accepted Submission(s): 25791

点击打开题目链接

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
非常简单基础的题目。

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<string>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#define INF 0x3f3f3f3f
#define MAXN 102

using namespace std;

int M,N;
struct room
{
int J;  ///javabean
int F;  ///food
double p; ///food/javabean
}r[100000];
bool cmp(struct room r1,struct room r2) ///按价格从小到大排序
{
return r1.p<r2.p;
}
int main()
{
while(~scanf("%d%d",&M,&N))
{
if(M == -1 && N == -1)
break;
for(int i = 0; i < N; i++)
{
scanf("%d%d",&r[i].J,&r[i].F);
r[i].p = (r[i].F*1.0)/r[i].J;
}
sort(r,r+N,cmp);
double ans = 0;
for(int i = 0; i < N; i++)
{
if(M >= r[i].F)
{
ans += r[i].J;
M = M - r[i].F;
}
else
{
ans += ((M*1.0)/r[i].F)*r[i].J;
M = 0;
}
}
printf("%.3lf\n",ans);
}
return 0;
}