杨柳之，象征着柔美与坚韧；天堑无涯，代表了难以逾越的障碍与无限可能。在这个世界里，有一种独特的书信系统，它跨越了千山万水，穿越了世俗与 sanitize.
每当 someone 推开 letterbox 的瞬间，总会有一缕杨柳之香飘入室内。那些装满故事的信纸，在岁月的打磨下愈发泛黄，却依然散发着温暖的气息。这些信都是通过某种神秘的方式传递的，仿佛上天派来的使者。
传说，远古时期有一段恩仇往事，只有 through the 杨柳之才能沟通天地，跨越千山万水。后来，这段故事被一代代人口耳相传，成为江湖上的传奇。
杨柳之不仅仅是一种象征，更是一种连接的力量。它让散落的信件得以传递，让远隔重洋的情谊得以延续。在这个信息爆炸的时代，书信系统反而显得弥足珍贵。
Okay, so I have this problem where I need to figure out how many cars can pass through an intersection in one hour. The traffic light at the intersection is green for 40 seconds, yellow for 5 seconds, and red for 65 seconds. Each car takes about 4 seconds to pass through the intersection once it’s clear.
Hmm, okay, let me break this down step by step. First off, I know that in one hour, which is 3600 seconds, the traffic light cycle repeats over and over again. So each full cycle of green, yellow, red takes a certain amount of time, and within each cycle, cars can pass through during the green phase.
So let’s calculate the duration of one complete traffic light cycle. Green is 40 seconds, yellow is 5, and red is 65. Adding those together: 40 + 5 + 65 equals… hmm, 40 plus 65 is 105, plus 5 makes 110 seconds. So each full cycle is 110 seconds.
Now, in each green phase, cars can pass through one after another. Each car takes about 4 seconds to clear the intersection. But wait, how many cars can pass during that green light? Since the green lasts for 40 seconds, and each car takes 4 seconds, I might think it’s simply dividing 40 by 4, which would give me 10 cars per green phase.
But hold on a second—is there any waiting time between cars or does each car start passing immediately after the previous one finishes? If we assume that cars are spaced closely enough and can pass without stopping, then yes, it should be possible to have 10 cars passing during each green light. But in reality, you might need some space between them, but since the problem doesn’t specify any gaps, I think we can go with 40 divided by 4, which is 10 cars per green phase.
So if there are 10 cars passing every green period of 40 seconds, and each green-red cycle repeats every 110 seconds, then how many cycles happen in one hour? Let me compute that.
First, the number of cycles in an hour: total time is 3600 seconds, divided by cycle duration of 110 seconds per cycle.
So 3600 / 110 equals approximately… let’s see, 110 goes into 3600 how many times? Well, 110 * 32 = 3520, which is less than 3600. Then the remainder is 80 seconds. So that would be about 32 full cycles in an hour.
Wait a minute, but actually, it’s not exactly precise because after 32 cycles, we have only 80 seconds left, which isn’t enough for another full cycle (since each cycle is 110 seconds). Therefore, the number of complete cycles within one hour would be 32, with some remaining time.
But hold on—is that how it works? Because traffic lights don’t necessarily reset exactly at the end of an hour; they just keep cycling through green, yellow, red continuously. So in reality, during the last partial cycle, we might get a little bit more cars passing if there’s any residual green time left after 32 cycles.
But maybe for simplicity, since it’s one hour total, and each cycle is 110 seconds, perhaps they just want us to calculate based on the number of full cycles within that hour. So let me confirm:
Total time = 3600 seconds
Cycle duration = 110 seconds
Number of cycles = floor(3600 / 110)
Calculating 3600 divided by 110: 110*32=3520, as above. So after 32 cycles (3520 seconds), we have 80 seconds left.
In those remaining 80 seconds, does the light stay green for a full 40 seconds? Or is it already in red?
Wait, no—the cycle is green (40), yellow (5), red (65). So after each full cycle of 110 seconds, the pattern repeats. Therefore, if we have 32 cycles, that brings us to 3520 seconds, and then from there, in the remaining 80 seconds, how much time is left?
In those 80 seconds, the traffic light would go through green (40) and yellow (5), making it a total of 45 seconds. But wait: 80 – 40 = 40; so after green, you have another 40 seconds remaining? No, actually, in reality, each cycle is fixed.
Wait, perhaps I need to model this as the light cycles through green, yellow, red repeatedly without overlap. So within one hour, how much time does it spend in green?
Another approach: instead of calculating number of cycles, calculate total green time in an hour and then see how many cars can pass during that green time.
That might be a better way because traffic lights don’t necessarily have fixed numbers of cycles if the total time is not a multiple of cycle duration. So maybe we should compute the fraction of the total hour spent in green light, multiply by 60 minutes to get it in seconds, then divide by how long each car takes.
Yes, that makes sense because regardless of how many full cycles there are, the amount of time the light is green over an hour can be calculated precisely.
So let’s try this alternative approach:
Total cycle duration: 40 (green) + 5 (yellow) + 65 (red) = 110 seconds
Time spent in green per cycle: 40 seconds
Fraction of time in green: 40 / 110, which simplifies to 4/11.
Therefore, total green time in one hour is (4/11) * 3600 seconds.
Calculating that:
(4/11)*3600 = (4*3600)/11
Compute numerator: 4*3600=14400
Divide by 11: 14400 /11 ≈ 1309.09 seconds.
So approximately, about 1309.09 seconds in green per hour.
Since each car takes 4 seconds to pass through, the number of cars passing during green time is total green time divided by time per car.
Number of cars = 1309.09 / 4 ≈ 327.27
But since we can’t have a fraction of a car, we need to take the integer part, which would be 327 cars.
Wait but is that accurate? Because in reality, as each green phase allows some number of cars, and with overlapping cycles…
Hmm, actually, no—it’s better to think per cycle. Wait, I’m confused now because this approach gives approximately 327 cars per hour, while the previous method gave about 10 cars per 40-second green phase multiplied by (number of times that happens in an hour). So let me see.
Earlier, I thought number of cycles is floor(3600 / 110) =32 cycles, each giving 10 cars, so 32*10=320. Then, the remaining time is 80 seconds—so during that time, how much green do we have?
Wait in those last 80 seconds: the light would cycle through green (40s), yellow (5s), red (65s). But with only 80 seconds left after 32 cycles.
So starting from a full cycle, each cycle is green, yellow, red. So if we have 80 seconds remaining after 32 cycles, the light would be in green for the first 40s of that last partial cycle, then yellow for next 5s, making it 45s passed, so total time elapsed now is 32*110 + 45 = 3520 +45=3565 seconds. But we only have 80 seconds left in the hour.
Wait no—no, after cycle 32 (3520 seconds), the light turns green again for another 40s.
But we only have 80 seconds left in the hour, so if it starts a new cycle of green at 3520 +40=3560, which is less than 3600. So during this last partial cycle:
From second 3520 to 3560: green (40s)
From 3560 to 3565: yellow
From 3565 to 3565+65=3630, which is beyond the hour.
So within the remaining time after cycle 32, there’s only 80 seconds left. Therefore, during this last partial green phase:
The light is green from 3520 to 3560 (40s), yellow at 3560-3565(5s). So within the remaining time of 80 seconds after cycle 32’s red phase—wait, no. Wait, actually, each full cycle is green, yellow, red.
Therefore, the state at second 3520 would be starting a new cycle: so it’s green from 3520 to 3560 (40s), then yellow from 3560-3565, and red from 3565 onwards. But since we only have till 3600 seconds:
So in the last partial cycle: Green for first 40s, yellow next 5s, red from there.
But wait, if total time elapsed is already 3520 (cycle 32’s end), and the hour ends at 3600. So how much green do we have?
From 3520 to 3600: that’s 80 seconds.
So starting from second 3520, which is a green phase:
Green duration for this partial cycle would be min(40, 80) =40s. Then yellow follows if there are remaining time after that.
Wait no—the light cycles through phases regardless of the time left—it’s just fixed: so from second 3520 to 3560 is green (40s), then at 3560-3565 yellow, and red until 3630.
But our observation ends at 3600. So during the last partial cycle:
From 3520 -3560: green
Then from 3560 to 3565: yellow
At 3565 onwards, it’s red until 3630.
But since our observation ends at 3600, the time between 3565 and 3600 is in red phase.
Therefore, during this last partial cycle:
From 3520 to 3560: green (40s)
So total green time over entire hour is cycles * green_time + residual_green
Which would be 32*40 + 40=1320 seconds.
Wait wait: that can’t be right because earlier calculation gave about 1309.09, but this gives 32*40+40=1320? But in reality, after cycle 32, which ends at 3520:
Cycle 33 is green for 40s (to 3560), yellow next 5s (to 3565). Then red from 3565 to 3630. So in the hour up to 3600, after cycle 32, only 80 seconds are left.
Therefore green phase at this last partial cycle is 40s (from 3520-3560), then yellow for next 5s (to 3565). So total green time in the hour would be:
Total cycles’ green: 32*40=1280
Plus residual green: 40
Total green time: 1280 +40=1320 seconds.
Which is more than previous calculation of ~1309.09. Hmm, that suggests maybe inaccuracy in the initial approach.
Wait why? Because:
(4/11)*3600 ≈ (4*3600)/11≈ 14400/11≈ 1309.09
But here we have residual green time as 40, but it’s not a fraction of the total.
Wait perhaps because when you compute over an exact hour, in reality:
Total cycles: 32 full cycles (each 110s), which is 3520 seconds
Then partial cycle:
From 3520 to 3600: 80 seconds
So within this last 80s, the light spends first 40s in green, then yellow for next 5s (totaling 45s), and remaining 35s are red.
Therefore total green time is:
32 cycles *40 + residual green=1600+40=1640? Wait no. No—each full cycle only has one green phase of 40s, so each cycle contributes 40s to green regardless.
Wait no: the first method was thinking in terms per cycle.
But if we have n cycles, that’s 110*n seconds
Then residual time is T -110*n=80s
In this residual time, how much green phase occurs?
So at the end of each full cycle (cycle k), the light turns green again for next 40s.
Therefore in residual time after n cycles:
If residual time >=40s: Then we have a partial green, yellow and red as before; else no green.
Wait so here residual is 80 seconds. So first 40 are green, then next 5 are yellow, rest 35 are red.
So total green in the hour= (n +1)*40 – time after that?
No:
Total green over all cycles: n *40
Plus any additional green from residual phase: min(residual_time,40)
But wait no. Each cycle is fixed: So for example, if you have partial cycles.
Wait perhaps a better approach is to compute the total number of full cycles and
