FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43381 Accepted Submission(s):
14499
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.
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.
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.
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
贪心题：题目的意思是FatMouse有m磅的cat food，它想用这些cat food去换JavaBean，然后有n个房间，每个房间有相应的JavaBean和需要多少cat food。问m磅cat food最多可以换得多少JavaBean？
首先要将这些数据按照性价比（即JavaBean除以cat food 注意性价比有可能是小数）从大到小排序，然后问题就迎刃而解了。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<iomanip>
using namespace std;
#define N 1005 struct Trade{
int J, F; //J代表JavaBeans，F代表cat food
double price; //性价比
}trade[N]; int cmp(Trade a,Trade b)
{
return a.price>b.price;
}
int main()
{
double maximum;
int m, n, i;
while(cin>>m>>n)
{
maximum = ;
if(m==- && n==-)
break;
maximum = ;
for(i=; i<n; i++)
{
cin>>trade[i].J>>trade[i].F;
trade[i].price = trade[i].J*1.0/trade[i].F;
}
sort(trade,trade+n,cmp);
for(i=; i<n; i++)
{
if(m == )
break;
else if(trade[i].F<=m)
{
maximum += trade[i].J;
m -= trade[i].F;
}
else
{
maximum += m*trade[i].price;
m = ;
}
}
printf("%.3lf\n",maximum);
}
return ;
}