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DOS批处理 & 脚本技术（批处理室） » Problem: How many times do the three hands of a clock coincide in a day?

Floor16　wxcute	Posted 2009-01-13 12:45
中级用户　Posts 211　Credits 458
According to everyone's understanding, the ideal clock should overlap 23 times. 0-24 contains 25 whole hours, 24 hours; Between 11-13, there is only the 12 o'clock overlap between the two whole hours, and similarly, between 23-24, there is only 24; 0-24 is equivalent to one more second in a whole day, and just this one second is the overlap, so it is 22+1=23 times.

Floor17　zh159	Posted 2009-01-13 18:16
金牌会员　Posts 1,467　Credits 3,687
Time is continuous like flowing water, not in jumps, so the ideal clock should be like what the 8th floor said

Floor18　linee	Posted 2009-01-14 00:13
初级用户　Posts 49　Credits 94
(Found online, didn't study the proof and don't understand) Only twice Assume the angular velocity of the hour hand is ω (ω = π/6 per hour), then the angular velocity of the minute hand is 12ω, and the angular velocity of the second hand is 72ω. Let t be the time when the minute hand and the hour hand coincide again, then 12ωt - ωt = 2π, t = 12/11 hours, converted to hours, minutes and seconds is 1 hour 5 minutes 27.3 seconds. Obviously, the second hand does not coincide with the hour hand and the minute hand. Similarly, it can be calculated that the second hand cannot coincide with them in the other 10 times when the minute hand and the hour hand coincide. Only at exactly 12 o'clock and 0 o'clock do they coincide. Proof: Treat the hour hand as stationary, and examine the relative speeds of the minute hand and the second hand with respect to it: Take 12 hours as the time unit "1", and "circles/12 hours" as the speed unit, then the speed of the minute hand is 11, and the speed of the second hand is 719. Since 11 and 719 are coprime, let 12 hours / (11719) be the time unit Δ, then the minute hand coincides with the hour hand if and only if t = 719kΔ, k ∈ Z The second hand coincides with the hour hand if and only if t = 11jΔ, j ∈ Z And the least common multiple of 719 and 11 is 11719, so if the three hands coincide at t = 0, then the next time the three hands coincide must be at t = 11719Δ, that is, t = 12 o'clock.

Floor19　slore	Posted 2009-01-14 10:25
铂金会员　Posts 2,478　Credits 5,212
It is said that if you calculate by angle, you won't get my result. This problem doesn't say the way the needle rotates. There are clocks that always run, and most of the designed needles still jump. The minimum swing amplitude is 1 second... This kind of needle is not in uniform circular motion... The above proof is not applicable.

Floor20　terse	Posted 2009-01-14 16:41
银牌会员　Posts 946　Credits 2,404
12-hour overlap @echo off for /l %%i in (0 1 11) do ( setlocal set/a n=%%i60 call:lp %%i endlocal ) pause&goto :eof :lp set/a d=%d%%n:~,1%%%11,m=%d%%n:~,1%/11 if "%i%"=="0" set i= set i=%i%%m% set n=%n:~1% if defined n goto lp set/a d=%160-(11i) :lp1 set/a m=d10/11,w+=1 if "%j%"=="0" set j= if %w% lss 5 set j=%j%%m%&set/a d=d10%%11&goto lp1 set/a j=6,h+=%1+i/60,i%%=60 set h=0%h%&set i=0%i%&set j=0%j% echo %h:~-2%:%i:~-2%:%j:~-5,2%.%j:~-3,2% goto :eof [ Last edited by terse on 2009-1-15 at 11:17 ]

Floor21　linee	Posted 2009-01-14 23:41
初级用户　Posts 49　Credits 94
Originally posted by terse at 2009-1-14 16:41: 12-hour coincidence @echo off for /l %%i in (0 1 11) do ( setlocal set/a n=%%i*60 call:lp %%i endlocal ) pause&goto :eof :lp set/a d=%d%%n:~,1%%%11,m=%d%%n:~,1%/11 ... The very version of the moderator can't understand this. Briefly introduce the algorithm.

Floor22　terse	Posted 2009-01-15 02:40
银牌会员　Posts 946　Credits 2,404
Originally posted by linee at 2009-1-14 23:41: The super moderator can't understand this. Simply introduce the algorithm. My thinking is like this. Let's discuss whether it's correct or not. Suppose the surface has 60 scales. One hour is 5 scales. x scales is x/5 scales after 0 o'clock. Similarly, y minutes is y/60 hours. When reaching the next coincidence point, the two scales are the same after 0 o'clock. That is, x/5 hours - y/60 hours is an integer. Because the minute hand goes one more circle, so it keeps increasing by 1. Because the scales are the same, so I think the result of X/5 - Y/60 should be increasing from 0 to 11. Here, x and y scales are equal, that is, X/5 - X/60. Just wrote a batch to derive another problem from this question: That is, at what scales the hour hand and the minute hand can be swapped and are reasonable, for example: 07:43:13.00 07:48:15.10 07:53:17.20 07:58:19.29 08:03:21.39 08:08:23.49 08:13:25.59 08:18:27.69 08:23:29.79 08:28:31.88 08:33:33.98 08:38:36.07

Floor23　linee	Posted 2009-01-15 09:28
初级用户　Posts 49　Credits 94
Originally posted by terse at 2009-1-15 02:40: My train of thought is like this. Let's discuss whether it's correct or not. Suppose the surface has 60 scales. One hour is 5 scales. x scales is x/5 scales after 0 o'clock. Similarly, y minutes is y/60 hours. When reaching the next coincidence point... It should be correct, very similar to my calculation (I calculate it by the fact that the hour hand and minute hand coincide once every 12/11 hours). As follows: Why is there no picture? Forget it, just explain in text. Your calculation is very similar to mine (provided earlier). There is only one abnormal moment, which is at 09:49. Your calculation is 09:49:54.540, and what I provided earlier is 09:49:05.45. There's a big difference between 54 and 05. It's a bit strange. Let's analyze your code first. [ Last edited by linee on 2009-1-15 at 09:54 ]

Floor24　terse	Posted 2009-01-15 11:18
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Originally posted by linee at 2009-1-15 09:28: ....One is 54 and the other is 05 The bit position was wrong, has been corrected

Floor25　linee	Posted 2009-01-15 15:10
初级用户　Posts 49　Credits 94
Originally posted by terse at 2009-1-15 02:40: My thinking is like this. Let's discuss whether it's correct or not. Assume the surface has 60 scales. One hour is 5 scales. x scales is x/5 scales after 0 o'clock. Similarly, y minutes is y/60 hours. When reaching the next coincidence point... After analysis, the result of yours should be equivalent to coinciding once every 12/11 hours. Try modifying your code as follows, hope you don't mind. @echo off for /l %%i in (0 1 11) do ( setlocal set/a n=%%i60 call:lp %%i endlocal ) pause&goto :eof :lp set/a d=%n%%%11,i=%n%/11 set/a j=d10000/11 set/a j*=6,h=%1+i/60,i%%=60 set h=0%h%&set i=0%i%&set j=0%j% echo %h:~-2%:%i:~-2%:%j:~-5,2%.%j:~-3,2% goto :eof

Floor26　terse	Posted 2009-01-15 15:26
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I actually solved the above program. You simplified a lot, which is very good. Let's discuss again at what scales the hour hand and minute hand can be swapped and be in a reasonable situation. Among them, the coincidence situation has been prompted. @echo off for /l %%i in (0 1 11) do ( for /l %%j in (0 1 11) do ( setlocal enabledelayedexpansion set/a n=^(%%j12+%%i^)60 set/a d=n%%143,i=n/143,j=d10000/143,j=6,h=%%i+i/60,i%%=60 set h=0!h!&set i=00!i!&set j=00!j! if %%i equ %%j (echo The hands coincide echo !h:~-2!:!i:~-2!:!j:~-5,2!.!j:~-3,2! ) else echo !h:~-2!:!i:~-2!:!j:~-5,2!.!j:~-3,2! endlocal )) pause&goto :eof The code for coincidence is placed in the for loop for higher efficiency. The previous call was because of considering decimals. Now it's directly okay. @echo off for /l %%i in (0 1 11) do ( setlocal enabledelayedexpansion set/a n=%%i60,d=n%%11,i=n/11,j=d10000/11,j=6,h=%%i+i/60,i%%=60 set h=0!h!&set i=00!i!&set j=00!j! echo !h:~-2!:!i:~-2!:!j:~-5,2!.!j:~-3,2! endlocal ) pause [ Last edited by terse on 2009-1-18 at 14:51* ]

Floor27　linee	Posted 2009-01-15 17:26
初级用户　Posts 49　Credits 94
What does "hour hand and minute hand exchange" mean?

Floor28　terse	Posted 2009-01-15 17:33
银牌会员　Posts 946　Credits 2,404
Originally posted by linee at 2009-1-15 17:26: What does "swap the positions of the hour hand and minute hand" mean? For example, when they overlap, swapping the positions of the hour hand and minute hand must be reasonable. For example, at 3 o'clock, swapping the positions is obviously unreasonable because when the hour hand is at 12 o'clock, the minute hand will never be at 3 o'clock.

Floor29　linee	Posted 2009-01-16 23:03
初级用户　Posts 49　Credits 94
Originally posted by terse at 2009-1-15 17:33: For example, when the time coincides, swapping the positions of the hour and minute hands is definitely reasonable. For example, at 3 o'clock, swapping the positions is obviously unreasonable because when the hour hand is at 12 o'clock, the minute hand will never be at 3 o'clock. A bit difficult to understand, don't want to study anymore.

Floor30　linee	Posted 2009-01-16 23:06
初级用户　Posts 49　Credits 94
Let me post and see how it goes, Ashamed, I regret not having read enough books when I need them. I always feel I can't express myself clearly. I've tried my best, just make do with it. Welcome to point out mistakes. :: Clock three-hand coincidence problem: :: :: Idea: First find the position s1 where the hour hand and minute hand coincide, then find the position s2 of the second hand at this time, :: If s2 = s1, then the three hands coincide. Compare from 0 o'clock to 24 o'clock. :: :: The first coincidence of the hour hand and minute hand occurs between 1 o'clock and 2 o'clock. Let the time at this moment be t. :: The angular velocity of the hour hand is w. At this time, the hour hand travels wt. The angular velocity of the minute hand is 12 times that of the hour hand. :: There is that the minute hand travels 12wt. At this time, the minute hand has traveled one full circle (12w) more than the hour hand. So there is: :: 12wt - wt = 12w, t = 12/11 :: At this time, the position of the hour hand wt = w * 12/11 = (1/11) * 12w, which is equivalent to 1/11 circle, can be taken :: s1 = 1/11 :: Also, the angular velocity of the second hand is 720 times that of the hour hand. At this time, the second hand travels 720wt. The position of the second hand :: 720wt = 720w * 12/11 = (65 + 5/11) * 12w :: Which is equivalent to 65 and 5/11 circles, can be taken :: s2 = 5/11 :: Compare s1 is not equal to s2, knowing that the three hands do not coincide at this time. Because cmd does not support floating points, :: The comparison program of s1 and s2 is changed to take the remainder of 11 and then compare. :: @echo off&setlocal enabledelayedexpansion for /l %%i in (0,1,22) do ( set/a s1=%%i %%11,s2=%%i5%%11,i=%%i+1 set/a h=%%i12/11,m=%%i60/11%%60,s=%%i300/11%%60,sh=%%i300%%11,sh=100/11 for %%j in (h m s sh i) do if !%%j! lss 10 set %%j=0!%%j! set /p=Hour-minute hand coincidence_!i! !h!:!m!:!s!.!sh! <nul if !s1! equ !s2! (set/a j+=1&echo Three-hand coincidence_!j!) else echo. ) pause [ Last edited by linee on 2009-1-16 at 23:14 ]

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