FatMouse' Trade

最新推荐文章于 2021-02-23 22:53:14 发布

原创最新推荐文章于 2021-02-23 22:53:14 发布 · 74 阅读

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通过贪心算法解决FatMouse使用猫粮换取JavaBeans的最大数量问题。输入包含多个测试案例，每个案例给出不同房间JavaBeans和所需猫粮的数量及比例，输出为FatMouse能获取的最大JavaBeans数量。

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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

Sample Output

13.333
31.500

这两天碰到不少贪心题了，这道也是一道贪心，用你自己有的猫食去有比例换取猫食，看你怎么能够换取最多的猫食，贪心问题的排序比较重要，把收入和付出的比例排序，然后从大到小换取。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int a,b;
double c;
}s[1010];
int cmp(node x,node y)
{
return x.c>y.c;
}
int main()
{
int n,m;
while(~scanf("%d%d",&m,&n))
{
if(m==-1&&n==-1)
break;
for(int i=0;i<n;i++)
{
scanf("%d%d",&s[i].a,&s[i].b);
s[i].c=(s[i].a*1.0)/s[i].b;
}
sort(s,s+n,cmp);
int i=0;
double sum=0;
while(m>0&&i<n)
{
if(m>=s[i].b)
{
sum+=s[i].a;
m-=s[i].b;
}
else
{
sum+=(m/(s[i].b*1.0))*s[i].a;
break;
}
i++;
}
printf("%.3lf\n",sum);
}
return 0;
}