杭电 FatMouse' Trade

问题地址

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6572 Accepted Submission(s): 1940

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

这道题有点坑，没有告诉你一定可以将M多的猫食全部换光，如果用while做的话，要判断是否还可以进行交换。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX_N 1005
#define MIN(a, b)   (a < b)? a: b
#define MAX(a, b)   (a > b)? a: b
using namespace std;

struct node {
    double f, j;
    double aver;
} r[MAX_N];

bool cmp(node a, node b) {
    return a.aver > b.aver;
}
int main() {
    int m, n;
    while (scanf("%d %d", &m, &n) != EOF) {
        if (m == -1 && n == -1) break;
        for (int i = 0; i < n; i++) {
            scanf("%lf %lf", &r[i].f, &r[i].j);
            r[i].aver = r[i].f / r[i].j;
        }
        double val = 0;
        sort(r, r + n, cmp);
        for (int i = 0; i < n; i++) {
            if (m == 0) break;
            if (m >= r[i].j) {
                m -= r[i].j;
                val += r[i].f;
            }
            else {
                val += r[i].aver * m;
                break;
            }
        }
        printf("%.3lf
", val);
    }
    return 0;
}