SDNU 1094

Description

There is an analog clock with two hands: anhour hand and a minute hand. The two hands form an angle. The angle is measuredas the smallest angle between the two hands. The angle between the two handshas a measure that is greater than or equal to 0 and less than or equal to 180degrees.

Given a sequence of five distinct timeswritten 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, thethird element of the sorted sequence of times in a nondecreasing order of theirassociated angles. Ties are broken in such a way that an earlier time precedesa 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. Thenumber of test cases (T) is given on the first line of the input file. Eachtest case is given on a single line, which contains a sequence of five distincttimes, where times are given in the format hh : mm and are separated by asingle 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. Thefollowing 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

这个题的意思是说：把每个时间在钟表上对应的时针和分针之间的夹角（小于180°）按顺序排列，把中间的一个输出。

这是WA的代码。。。

#include <iostream>#include <cstdio>#include <memory.h>#include <cmath>#include <algorithm>using namespace std;struct stuff{    int ho;    int mi;    int sign;}time[10];bool cmp(stuff a, stuff b){    if(a.sign > b.sign)        return true;    else if(a.sign == b.sign)    {        if(a.ho > b.ho)            return true;        else            return false;    }    else        return false;}int main(){    int n;    scanf("%d", &n);    while(n--)    {        memset(time, 0, sizeof(time));        int i;        for(i = 0; i < 5; i++)        {            scanf("%d:%d", &time[i].ho, &time[i].mi);            if(time[i].ho > 12)                time[i].sign = abs(30.0 * (time[i].ho - 12) - time[i].mi * 6.0 );            else                time[i].sign = abs(30.0 * time[i].ho - time[i].mi * 6.0);            if(time[i].sign > 180)                time[i].sign = 360 - time[i].sign;        }        sort(time, time + 5, cmp);        printf("%02d:%02d\n", time[2].ho, time[2].mi);    }    return 0;}

为什么WA呢？因为当分针从12开始走动时，时针也在慢慢的动（生活常识），所以正确的应该是：

#include <iostream>#include <cstdio>#include <memory.h>#include <cmath>#include <algorithm>using namespace std;struct stuff{    int ho;    int mi;    int sign;}time[10];bool cmp(stuff a, stuff b){    if(a.sign > b.sign)        return true;    else if(a.sign == b.sign)    {        if(a.ho > b.ho)            return true;        else            return false;    }    else        return false;}int main(){    int n;    scanf("%d", &n);    while(n--)    {        memset(time, 0, sizeof(time));        int i;        for(i = 0; i < 5; i++)        {            scanf("%d:%d", &time[i].ho, &time[i].mi);            if(time[i].ho > 12)                time[i].sign = abs(30.0 * (time[i].ho - 12) - time[i].mi * 6.0 + time[i].mi /2.0);            else                time[i].sign = abs(30.0 * time[i].ho - time[i].mi * 6.0 + time[i].mi /2.0);            if(time[i].sign > 180)                time[i].sign = 360 - time[i].sign;        }        sort(time, time + 5, cmp);        printf("%02d:%02d\n", time[2].ho, time[2].mi);    }    return 0;}

你能看出改动在哪吗？