"HDU 5387" Clock

Problem

Description

Give a time. (HH:MM:SS), should answer the angle between any of the Minute.hour.second hand
Notice that the answer must is not more than and not less than 0

Input

There is $T $$ (1\leq T \leq 10^4) $ test Cases
For each case,one line include the time

$0\leq Hh<24$,$0\leq Mm<60$,$0\leq ss<60$

Output

For each case,output there real number like A/b. (A and B are coprime). If it's an integer then just print It.describe the A Ngle between hour and Minute,hour and second hand,minute and second hand.

Sample Input

400:00:0006:00:0012:54:5504:40:00

Sample Output

0 0 0 180 180 0 1391/24 1379/24 1/2 100 140 120

Hint

Test instructions: The angle of the hour, minute, and second hand at a given point in time (fractional, angular)

Analysis: To be continued ....

#include <stdio.h>int t,h,m,s,tol,up[4],down[4],g;int gcd (int a,int b) {    return b?gcd (b,a%b): A;} int main () {    scanf ("%d", &t);    while (t--)    {        scanf ("%d:%d:%d", &h,&m,&s);        Tol=h*3600+m*60+s;        up[0]= (11*tol)% (43200);        down[0]=120;        up[1]= (719*tol)% (43200);        down[1]=120;        up[2]= (59*tol)% (3600);        down[2]=10;        for (int i=0; i<3; i++)        {            if (up[i]/down[i]>179) up[i]=down[i]*360-up[i];            G=GCD (Up[i],down[i]);            up[i]/=g;            down[i]/=g;            if (down[i]==1) printf ("%d", Up[i]>180?360-up[i]:up[i]);            else printf ("%d/%d", Up[i],down[i]);        }        printf ("\ n");    }    return 0;}

"HDU 5387" Clock