杭电OJ 题 1209 Clock 解题报告

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2985    Accepted Submission(s): 914

Problem Description

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

300:00 01:00 02:00 03:00 04:0006:05 07:10 03:00 21:00 12:5511:05 12:05 13:05 14:05 15:05

 

Sample Output

02:0021:0014:05

 

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注意时间的计算方法，时针是跟着分针动的，比如7：30的角度不是7点的角度和30分的角度只差，小时角度的计算还要加上分针的偏移量

还有当角度相同时，时间小的在前面

/************************   程序名:clock.cpp*   功能:ACM************************/#include <stdio.h>#include <string.h>void PubbleSort(double (*)[3], int);int main(){    int T;    char time[5][5], string[30], c;    double sequence[5][3];    int i, j, t;    double hour, min, degree;    scanf("%d", &T);    c=getchar();    while(T--) {        gets(string);        t = 0;        for(i = 0; i < 5; i++) {            for(j = 0; j < 5; j++) {                time[i][j] = string[t++];            }            t++;        }        for(i = 0; i < 5; i++) {            hour = (time[i][0]-'0')*10+time[i][1]-'0';            min = (time[i][3]-'0')*10+time[i][4]-'0';            sequence[i][2] = hour;            if(hour > 12) {                hour -= 12;            }            if((hour+min/60.0)*30 > min*6) {                degree = (hour+min/60.0)*30-min*6;            }            else {                degree = min*6-(hour+min/60.0)*30;            }            if(degree > 180) {                degree = 360 - degree;            }            sequence[i][0] = degree;            sequence[i][1] = i;        }        PubbleSort(sequence, 5);        i = sequence[2][1];        for(j = 0; j < 5; j++) {            printf("%c", time[i][j]);        }        printf("\n");    }    return 0;}void PubbleSort(double (*array)[3], int n){    int i, j;    double temp_1, temp_2, temp_3;    for(i = 0; i < n-1; i++) {        for(j = 0; j < n-i-1; j++) {            if(array[j][0] >= array[j+1][0]) {                if(array[j][0] == array[j+1][0]) {                    if(array[j][2] > array[j+1][2]) {                    temp_1 = array[j][0];                    temp_2 = array[j][1];                    temp_3 = array[j][2];                    array[j][0] = array[j+1][0];                    array[j][1] = array[j+1][1];                    array[j][2] = array[j+1][2];                    array[j+1][0] = temp_1;                    array[j+1][1] = temp_2;                    array[j+1][2] = temp_3;                    }                }                else {                    temp_1 = array[j][0];                    temp_2 = array[j][1];                    temp_3 = array[j][2];                    array[j][0] = array[j+1][0];                    array[j][1] = array[j+1][1];                    array[j][2] = array[j+1][2];                    array[j+1][0] = temp_1;                    array[j+1][1] = temp_2;                    array[j+1][2] = temp_3;                }            }        }    }    /*for(i = 0; i < 5; i++) {        printf("degree=%.0lf hour=%.0lf ", array[i][0], array[i][2]);    }    printf("\n");*/}