hdu FatMouse' Trade

最新推荐文章于 2020-12-04 15:13:07 发布

风吹过的烟花

最新推荐文章于 2020-12-04 15:13:07 发布

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分类专栏： ACM算法_贪心文章标签：贪心

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解决FatMouse如何用有限的猫粮换取最大数量的JavaBeans的问题。通过贪心算法选择性价比最高的交换方式。

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FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 1

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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

Sample Output

13.333
31.500

贪心，能买性价比高的，就买高的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int p,q;
    double v;
} s[1005];
int cmp(node x,node y)
{
    return x.v>y.v;
}
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
            break;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&s[i].p,&s[i].q);
            s[i].v=s[i].p*1.0/s[i].q;
        }
        sort(s,s+n,cmp);
        double ans=0;
        for(int i=0;i<n;i++)
        {
            if(m>=s[i].q)
            {
                ans+=s[i].p;
                m-=s[i].q;
            }
            else
            {
                ans+=s[i].p*1.0/s[i].q*m;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}