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# CLOCK:

Contest Link

Author: Tushar Kadam

Tester: Sangram Desai

Editorialist: Tushar Kadam

SIMPLE

# PROBLEM:

Given time in HH MM SS format and you have to find what will be time after hour hand moves by x degrees.

# EXPLANATION:

If hour hand moves by 1 degree then 2 minutes will happen in 12 hour clock. As x is integer and smallest change in x can be only 1 degree hence second hand will come to its original position whatever may be the value of x i.e SS field will not change. Now find if minutes to add in MM field is greater than 60 or not . If final MM field > 60 then add extra minutes in hour by converting minutes to hour.

Note : if x > 360 that means hour hand has moved by more than 360 degrees i.e several complete rotations have happened. so to solve one can do x = x%360 which will have same effect.

Possible testcases where you may get wrong answers:

input :

6

11 58 59 1

11 59 1 1

12 58 55 28

11 2 23 29

11 2 0 361

11 59 27 62

output:

12 0 59

12 1 1

1 54 55

12 0 23

11 4 0

2 3 27

# AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution Java Solution

Tester's solution Python Solution.

This question is marked "community wiki".

asked 09 Sep, 14:22

5★cis_pie
355
accept rate: 12%

0★admin ♦♦
19.3k348495534

 1 The preliminary step here is to find out what $1$ degree motion of the hour hand means in terms of time. 0ne full rotation of the hour hand is $12$ hours or $12\times 60 = 720$ minutes, and conveniently we have $720/360 = 2$ exactly; giving $2$ minutes per degree. If, instead of degrees, the question had asked the time after the hour hand is moved by $x$ gradians, which have $400$ for a full turn, we would have had to go down to an effect in seconds, with $720\times 60 = 43200$ second in a full turn of the hour hand and thus $108$ seconds per gradian. answered 13 Sep, 02:43 5★joffan 418●6 accept rate: 9% @joffan Thanks for your suggestion about use of gradians in problem . But I wanted to design an easy problem So I deliberately made a special case in which that it turns out that 1 degree corresponds to 2 minutes. So bonus question : x can be negative (13 Sep, 23:43) cis_pie5★ 1 Oh sure, I wasn't complaining at all about the use of degrees. Just pointing out where the 1 degree = 2 minutes comes from. And my alternative example was gradians not radians - they would be much more difficult since a full turn in radians is $2\pi$. PS: re bonus question: my code already worked correctly. :-) (13 Sep, 23:51) joffan5★ Oh my bad! Edited! And I just saw your code. (14 Sep, 00:09) cis_pie5★

# include<time.h>

long t,x[100000]; int hr[100000],min[100000],sec[100000]; void calc(){ long m=0,min_angle,hr_angle,sec_angle; while(m<t){ if(x[m]%30="=0){" x[m]="x[m]/30;" if(x[m]="">12){ while(x[m]>12){ x[m]=x[m]-12; } } hr[m]=x[m]+hr[m]; if(hr[m]>12){ hr[m]=hr[m]-12; } } else{ if(x[m]<30){ min_angle=x[m]; hr_angle=0; } else{ min_angle=x[m]%30; hr_angle=x[m]/30; } if(hr_angle>12){ while(hr_angle>12){ hr_angle=hr_angle-12; } } hr[m]=hr[m]+hr_angle; if((min_angle2+min[m])>=60){ hr[m]=hr[m]+1; min[m]=(min[m]+(min_angle2))-60; } else{ min[m]=min[m]+min_angle*2; } if(hr[m]>12){ hr[m]=hr[m]-12; } } m++; } } int main(void) { scanf("%lu",&t); for(long i=0;i<t;i++){ scanf("%d%d%d%li",&hr[i],&min[i],&sec[i],&x[i]); } calc(); for(long m=0;m<t;m++){ printf("%d %d %d\n",hr[m],min[m],sec[m]); } return 0; }

answered 12 Sep, 21:22

0★girya
1
accept rate: 0%

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question asked: 09 Sep, 14:22

question was seen: 361 times

last updated: 14 Sep, 18:07