杭电oj 1009

先看题目

FatMouse' Trade

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

Total Submission(s): 57721 Accepted Submission(s): 19329



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

[align=left]Sample Output[/align]

13.333
31.500

题目的意思是FatMouse用CatFood交换JavaBean（JavaBean乱入，哈哈，博主有空写一个关于javaBean的博客），FatMouse拥有M的CatFood，然后要交换最多的Javabean。
很明显是贪心算法，我们对JavaBean/CatFood进行排序，问题就很容易解决了，还有要注意的是，当M=0和N=0时的情况
下面贴上我的代码

#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
struct Rate
{
double javaBean;
double catFood;
double rat;
}rate[1000];
bool cmp(const Rate &a,const Rate &b)
{
if(a.rat>b.rat)
return true;
else
return false;
}

int main()
{
int M,N;
int i;
double sum;
while(cin>>M>>N&&M!=-1&&N!=-1)
{
sum=0;
if(N!=0)
{
for(i=0;i<N;i++){
cin>>rate[i].javaBean>>rate[i].catFood;
rate[i].rat=rate[i].javaBean/rate[i].catFood;
}
sort(rate,rate+N,cmp);
i=0;
while(M>0)
{
if(M>rate[i].catFood)
{
M-=rate[i].catFood;
sum+=rate[i].javaBean;
}
else
{
sum+=rate[i].rat*M;
break;
}
i++;
}
}

cout<<setprecision(3)<<std::fixed<<sum<<endl;
}
return 0;
}

ac了