暑假楼下第三场练习赛——A - FatMouse' Trade（贪心）

最新推荐文章于 2025-06-13 13:59:32 发布

转载最新推荐文章于 2025-06-13 13:59:32 发布 · 165 阅读

FatMouse准备了M磅猫粮，计划与守卫仓库的猫咪们进行交换，以获取其最爱的食物——Java豆。仓库中有N个房间，每个房间内的Java豆数量与所需猫粮数量各不相同。本题要求通过合理分配猫粮来最大化获取的Java豆总量。

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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 Outpu

13.333 31.500

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
struct T
{
int f;
int j;
double p;
} q[1001];
bool cmp(T a,T b)
{
return a.p>b.p;
}

int main()
{
int m,n,i,j;
while(cin>>m>>n)
{

int k=0;
if(n==-1&&m==-1)
break;
for(i=0; i<n; i++)
{
scanf("%d %d",&q[i].j,&q[i].f);
q[i].p=(double)q[i].j/q[i].f;

}
sort(q,q+n,cmp);
double sum=0;
for(i=0; i<n; i++)
{

if(m>q[i].f)
{
sum+=q[i].j;
m-=q[i].f;
}
else
{
sum+=q[i].p*m;
m=0;
break;
}
}
printf("%.3lf\n",sum);
}
return 0;
}