ZOJ 2680 Clock（）数学

大家好，又见面了，我是全栈君

主题链接：There is an analog clock with two hands: an hour hand and a minute hand. The two hands form an angle. The angle is measured as the smallest angle between the two hands. The angle between the two hands has a measure that is greater than or equal to 0 and less than or equal to 180 degrees.

Given a sequence of five distinct times written in the format hh : mm , where hh are two digits representing full hours (00 <= hh <= 23) and mm are two digits representing minutes (00 <= mm <= 59) , you are to write a program that finds the median, that is, the third element of the sorted sequence of times in a nondecreasing order of their associated angles. Ties are broken in such a way that an earlier time precedes a later time.

For example, suppose you are given a sequence (06:05, 07:10, 03:00, 21:00, 12:55) of times. Because the sorted sequence is (12:55, 03:00, 21:00, 06:05, 07:10), you are to report 21:00.

Input

The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. Each test case is given on a single line, which contains a sequence of five distinct times, where times are given in the format hh : mm and are separated by a single space.

Output

Print exactly one line for each test case. The line is to contain the median in the format hh : mm of the times given. The following shows sample input and output for three test cases.

Sample Input

3
00:00 01:00 02:00 03:00 04:00
06:05 07:10 03:00 21:00 12:55
11:05 12:05 13:05 14:05 15:05

Sample Output

02:00
21:00
14:05

Source: Asia 2003, Seoul (South Korea)

题意：

//给出 5 个时刻，按时钟的时针，分针夹角从小到大排序， //输出中间的时刻。

代码例如以下：

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
struct TIME
{
    int h;
    int m;
    int angle;
}a[7];

int cal(TIME TT)
{
    if(TT.h > 12)
    {
        TT.h-=12;
    }
    int tt = abs((TT.h*60 + TT.m) - TT.m*12);
    //原式为:TT.h*30+(TT.m/60)*30-a.m*6;
    if(tt > 360)
        tt = 720 - tt;
    return tt;
}
bool cmp(TIME A, TIME B)
{
    if(A.angle != B.angle)
    {
        return A.angle < B.angle;
    }
    else if(A.h != B.h)
    {
        return A.h < B.h;
    }
    else
        return A.m < B.m;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i = 0; i < 5; i++)
        {
            scanf("%d:%d",&a[i].h,&a[i].m);
            a[i].angle = cal(a[i]);
        }
        sort(a,a+5,cmp);
        printf("%02d:%02d\n",a[2].h,a[2].m);
    }
    return 0;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116756.html原文链接：https://javaforall.cn