hdu_1009 贪心

Posted 2021-11-05 dawn-bin

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105467    Accepted Submission(s): 36835

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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.

 

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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.

 

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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.

 

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Sample Input

5 3

7 2

4 3

5 2

20 3

25 18

24 15

15 10

-1 -1

 

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Sample Output

13.333

31.500

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经典贪心，老鼠要换取到最多的JavaBean，按照每个房间的j和f，算出比率，当然是j/f越大越好，按照比率排一下序。

#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

struct dat
{
    int j;
    int f;
    double sc;
} data[1000];

bool cmp(dat a, dat b)
{
    return a.sc>b.sc;
}

int main()
{
    int m,n;
    double ans;
    while(scanf("%d%d",&m,&n) && m!=-1 && n!=-1)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&data[i].j, &data[i].f);
            data[i].sc = (double)data[i].j/(double)data[i].f;
        }

        sort(data, data+n, cmp);

        ans = 0;
        for(int i=0; i<n; i++)
        {
            if(data[i].f<=m)
            {
                ans+=data[i].j;
                m-=data[i].f;
            }
            else
            {
                ans+=data[i].sc*(double)m;
                break;
            }

        }
        printf("%.3f\n", ans);
    }

    return 0;
}