题目描述
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.
输入
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.
输出
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.
样例输入
4 2
4 7
1 3
5 5
4 8
3 8
1 2
2 5
2 4
-1 -1
样例输出
2.286
2.500
思路:可以用一个结构体来存放第i个仓库的食物还有需要用于交换的猫的食物,以及二者之间交换的比例。定义一个double型变量result,用于最后输出。输入完数据后,将此结构体按照 j/f 的比例从大到小进行排序,然后从比例最大的那个仓库开始进行交换即可。如果m大于第i个仓库所需要交换时的猫粮,则令result加上该仓库用于交换的食物,并且令m - f [ i ],否则就加上该仓库的交换比例与剩下m的积。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct food{
double j;
double f;
double b;
}a[1000];
bool cmp(struct food a,struct food b){
if(a.b==b.b) return a.j>b.j;
else return a.b>b.b;
}
int main()
{
int n,i;
double m,result;
while(scanf("%lf%d",&m,&n)!=EOF){
if(m==-1&&n==-1)
break;
result=0;
for(i=0;i<n;i++){
cin >>a[i].j>>a[i].f;
a[i].b=a[i].j/a[i].f;
}
sort(a,a+n,cmp);
for(i=0;i<n;i++){
if(m<=0)
break;
if(m>=a[i].f)
result+=a[i].j;
else
result+=a[i].b*m;
m-=a[i].f;
}
printf("%.3lf\n",result);
}
return 0;
}