HDU 1009：FatMouse' Trade（简单贪心）

Posted 2020-09-13 claireyuancy

FatMouse‘ Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41982    Accepted Submission(s): 13962

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

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

 

Sample Output

13.333
31.500

  

题意就是老鼠用猫粮换鼠粮。

。

。（Orz）。。

求它最多能换多少。。一道贪心水题。

。

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<cmath>

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, n) for(int i=1; i<=n; i++)

using namespace std;

const int M = 1005;
int n, m;
double ans;
double t;

struct node
{
    double J;
    double F;
    double c;
}Q[M];

int cmp (node a, node b)
{
    return a.c > b.c;
}

int main()
{
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        ans = 0;
        t = (double) n;
        memset(Q, 0, sizeof(Q));
        if(n==-1 && m==-1)
            break;
        f1(i, m)
           {
               scanf("%lf%lf", &Q[i].J, &Q[i].F);
               Q[i].c = Q[i].J / Q[i].F;
           }
        sort(Q, Q+m, cmp);
        f1(i, m)
        {
            if( Q[i].F<=t )
                {
                    ans += Q[i].J;
                     t -= Q[i].F;
                }
            else
            {
                ans += t * Q[i].c;
                break;
            }
        }
        printf("%.3lf\n", ans);
    }

    return 0;
}