【九度OJ】题目1433：FatMouse 解题报告

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负雪明烛 2022-03-02 15:30:15 博主文章分类：算法 ©著作权

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【九度OJ】题目1433：FatMouse 解题报告

标签（空格分隔）：九度OJ

http://ac.jobdu.com/problem.php?pid=1433

题目描述：

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.

输入：

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.

输出：

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.

样例输入：

样例输出：

13.333
31.500

Ways

开始贪心算法的题。

这个题目的意思是，M的cat foode去每个room里换JavaBean，每个房间都是按等比例换。

可以证明，只有把rate=JavaBean/catFoode值最大的先换了，才会使最终换的的javaBean最多。

也就是相当于换性价比最高的东西，这样就会总获得量最大。

这就需要排序，我在做这个题的时候遇到一个问题，是自己考虑的不周到。刚开始没有把每个房间当做一个整体，这样在rate排序之后，rate和房间对应不上。把room作为整体就能解决了。

第二个问题还是没有把answer在每个循环中清零。

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

using namespace std;

struct room {
    float cat;
    float bean;
    float rate;
} rooms[1000];

bool cmp(room a, room b) {
    return a.rate > b.rate;
}

int main() {
    int m;
    int n;
    float answer;
    while (scanf("%d", &m) != EOF && m != -1) {
        answer = 0.0;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%f%f", &rooms[i].cat, &rooms[i].bean);
            rooms[i].rate = rooms[i].cat / rooms[i].bean;
        }
        sort(rooms, rooms + n, cmp);
        for (int i = 0; i < n ; i++) {
            if (m > rooms[i].bean) {
                answer += rooms[i].cat;
                m -= rooms[i].bean;
            } else {
                answer += (((float) m) / rooms[i].bean) * rooms[i].cat;
                break;
            }
        }
        printf("%.3f\n", answer);
    }
    return 0;
}