FatMouse’ Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20968 Accepted Submission(s): 6501
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
Recommend JGShining 一开始我没有想到用结构体来表示。程序快要敲完的时候发现记录比率的数组没法记录容纳量，无法F[i]* a。 于是我就在网上搜了有关的答案，用的是结构体。答案如下：
#include <iostream>
#include<algorithm>
#include <iomanip>
using namespace std;
const int MAXN=1010;
struct node
{
int J,F;
float ratio1;
}a[MAXN];
bool cmp(node a,node b)
{
return a.ratio1>b.ratio1;
}
int main()
{
int m,n,i=0,j;
float sum;
while (cin>>m>>n&&m!=-1)
{
sum=0;
i=0;
for (j=0;j<n;j++)
{
cin>>a[i].J>>a[i].F;
a[i].ratio1=(float)a[i].J/a[i].F;
i++;
}
sort(a,a+n,cmp);
for (i=0;i<n;i++)
{
if (m>=a[i].F)
{
sum+=a[i].J;
m-=a[i].F;
}
else
{
sum+=a[i].ratio1*m;
break;
}
}
cout<<fixed<<setprecision(3)<<sum<<endl;
}
return 0;
}
后来想了一下其实也可以不用结构体，就是在得到比率的同时对原来的数组进行排序，以达到计算是否超出容量的目的。
