第九天（2）：贪心算法

T1

FatMouse’ Trade
时间限制:1 秒
内存限制:128 兆
特殊判题:否

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

样例输出:
13.333
31.500

//
//  main.cpp
//  FatMouseTrade
//
//  Created by Apple on 2019/8/13.
//  Copyright © 2019 Apple_Lance. All rights reserved.
//

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

using namespace std;

struct goods{
    double j;
    double f;
    double s;
    
    bool operator < (const goods & A)const{
        return s > A.s;
    }
}buf[1000];

int main(int argc, const char * argv[]) {
    // insert code here...
    double m;
    int n;
    while (scanf("%lf %d", &m, &n) != EOF) {
        if(m == -1 && n == -1)
            break;
        for(int i = 0; i < n; i++){
            scanf("%lf%lf", &buf[i].j, &buf[i].f);
            buf[i].s = buf[i].j / buf[i].f;
        }
        sort(buf, buf + n);
        int idx = 0;
        double ans = 0;
        while(m > 0 && idx < n){
            if(m > buf[idx].f){
                ans += buf[idx].j;
                m -= buf[idx].f;
            }
            else{
                ans += buf[idx].j * m / buf[idx].f;
                m = 0;
            }
            idx ++;
            
        }
        printf("%.3lf\n", ans);
    }
    return 0;
}

T2

今年暑假不 AC
时间限制:1 秒
内存限制:128 兆
特殊判题:否

题目描述: “ 今 年 暑 假 不 AC ? ”“ 是 的 。 ”“ 那 你 干 什 么 呢 ? ”“ 看 世 界 杯 呀 , 笨蛋! ”“@#$%^&*%…”确实如此, 世界杯来了, 球迷的节日也来了, 估计很多ACMer 也会抛开电脑,奔向电视作为球迷,一定想看尽量多的完整的比赛,当然,作为新时代的好青年,你一定还会看一些其它的节目,比如新闻联播(永远不要忘记关心国家大事) 、非常 6+7、超级女生,以及王小丫的《开心辞典》等等,假设你已经知道了所有你喜欢看的电视节目的转播时间表, 你会合理安排吗? (目标是能看尽量多的完整节目)

输入: 输入数据包含多个测试实例,每个测试实例的第一行只有一个整数n(n<=100),表示你喜欢看的节目的总数,然后是 n 行数据,每行包括两个数据Ti_s,Ti_e (1<=i<=n),分别表示第 i 个节目的开始和结束时间,为了简化问题,每个时间都用一个正整数表示。n=0 表示输入结束,不做处理。
输出: 对于每个测试实例, 输出能完整看到的电视节目的个数, 每个测试实例的输出占一行。
样例输入:
12
1 3
3 4
0 7
3 8
15 19
15 20
10 15
8 18
6 12
5 10
4 14
2 9
0

样例输出:
5

//
//  main.cpp
//  Summer_AC
//
//  Created by Apple on 2019/8/13.
//  Copyright © 2019 Apple_Lance. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
struct program{
    int startTime;
    int endTime;
    bool operator < (const program & A)const{
        return endTime < A.endTime;
    }
}buf[1000];

int main(int argc, const char * argv[]) {
    // insert code here...
    int n;
    while(scanf("%d", &n) != EOF){
        if(n == 0)
            break;
        for(int i = 0;i<n;i++)
            scanf("%d%d", &buf[i].startTime, &buf[i].endTime);
        sort(buf, buf+n);
        int currentTime = 0, ans = 0;
        for(int i = 0;i<n;i++)
            if(currentTime <= buf[i].startTime){
                currentTime = buf[i].endTime;
                ans++;
            }
        printf("%d\n", ans);
    }
    return 0;
}