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August 6, 2021 00:33
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// source https://jsbin.com
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| <!DOCTYPE html> | |
| <html> | |
| <head> | |
| <meta charset="utf-8"> | |
| <meta name="viewport" content="width=device-width"> | |
| <title>JS Bin</title> | |
| </head> | |
| <body> | |
| <script id="jsbin-javascript"> | |
| /** | |
| * The goal is to make sure the last bike when there is only one bike left, | |
| * it can ride its full distance (100km / 1 (bike)). | |
| * When there are two bikes left, both of them could ride their full distance (100km / 2(bikes)) | |
| * This requires the second last bike to give away whatever the amount it has left to make up | |
| * the last bike to have 100km distance. Because the last two bikes must start at the same time when | |
| * both have 100km to go, and the second last bike must give its remaining fuel to the last bike ONLY | |
| * when R(n-1) + R(n) = 100km, where R is the function to represent the remaining fuel. Because they start | |
| * the same time, so we have: | |
| * 100km - R(n-1) = 100km - R(n) | |
| * => R(n) = R(n-1) | |
| * Replace R(n-1) with R(n): | |
| * R(n-1) + R(n) = 100km | |
| * => 2R(n) = 100km | |
| * => R(n) = R(n-1) = 50km | |
| * The same logics apply to any of the previous bikes back to the begining when we still have 50 bikes: | |
| * So the first leg when all 50 bikes start together, we need to stop at a distance where the remaining | |
| * fuel from one bike could be used to fill the rest 49 bikes. | |
| * S(1) = 100 / 50(bikes), where S is the distance to travel for the first leg. | |
| * Combine the above, we could derive the following formular | |
| * S(n) = 100/50 + 100/49 + 100/48 + ... + 100/i + ... + 100/3 + 100/2 + 100/1, where i is the ith leg. | |
| * Use the following iteration we get to know the sum. | |
| */ | |
| let sum = 0; | |
| for (let i = 50; i >=1; i--) { | |
| sum += 100/i; | |
| } | |
| console.log(sum) | |
| </script> | |
| <script id="jsbin-source-javascript" type="text/javascript">/** | |
| * The goal is to make sure the last bike when there is only one bike left, | |
| * it can ride its full distance (100km / 1 (bike)). | |
| * When there are two bikes left, both of them could ride their full distance (100km / 2(bikes)) | |
| * This requires the second last bike to give away whatever the amount it has left to make up | |
| * the last bike to have 100km distance. Because the last two bikes must start at the same time when | |
| * both have 100km to go, and the second last bike must give its remaining fuel to the last bike ONLY | |
| * when R(n-1) + R(n) = 100km, where R is the function to represent the remaining fuel. Because they start | |
| * the same time, so we have: | |
| * 100km - R(n-1) = 100km - R(n) | |
| * => R(n) = R(n-1) | |
| * Replace R(n-1) with R(n): | |
| * R(n-1) + R(n) = 100km | |
| * => 2R(n) = 100km | |
| * => R(n) = R(n-1) = 50km | |
| * The same logics apply to any of the previous bikes back to the begining when we still have 50 bikes: | |
| * So the first leg when all 50 bikes start together, we need to stop at a distance where the remaining | |
| * fuel from one bike could be used to fill the rest 49 bikes. | |
| * S(1) = 100 / 50(bikes), where S is the distance to travel for the first leg. | |
| * Combine the above, we could derive the following formular | |
| * S(n) = 100/50 + 100/49 + 100/48 + ... + 100/i + ... + 100/3 + 100/2 + 100/1, where i is the ith leg. | |
| * Use the following iteration we get to know the sum. | |
| */ | |
| let sum = 0; | |
| for (let i = 50; i >=1; i--) { | |
| sum += 100/i; | |
| } | |
| console.log(sum)</script></body> | |
| </html> |
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| /** | |
| * The goal is to make sure the last bike when there is only one bike left, | |
| * it can ride its full distance (100km / 1 (bike)). | |
| * When there are two bikes left, both of them could ride their full distance (100km / 2(bikes)) | |
| * This requires the second last bike to give away whatever the amount it has left to make up | |
| * the last bike to have 100km distance. Because the last two bikes must start at the same time when | |
| * both have 100km to go, and the second last bike must give its remaining fuel to the last bike ONLY | |
| * when R(n-1) + R(n) = 100km, where R is the function to represent the remaining fuel. Because they start | |
| * the same time, so we have: | |
| * 100km - R(n-1) = 100km - R(n) | |
| * => R(n) = R(n-1) | |
| * Replace R(n-1) with R(n): | |
| * R(n-1) + R(n) = 100km | |
| * => 2R(n) = 100km | |
| * => R(n) = R(n-1) = 50km | |
| * The same logics apply to any of the previous bikes back to the begining when we still have 50 bikes: | |
| * So the first leg when all 50 bikes start together, we need to stop at a distance where the remaining | |
| * fuel from one bike could be used to fill the rest 49 bikes. | |
| * S(1) = 100 / 50(bikes), where S is the distance to travel for the first leg. | |
| * Combine the above, we could derive the following formular | |
| * S(n) = 100/50 + 100/49 + 100/48 + ... + 100/i + ... + 100/3 + 100/2 + 100/1, where i is the ith leg. | |
| * Use the following iteration we get to know the sum. | |
| */ | |
| let sum = 0; | |
| for (let i = 50; i >=1; i--) { | |
| sum += 100/i; | |
| } | |
| console.log(sum) |
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