{"id":3544,"date":"2025-11-22T17:38:34","date_gmt":"2025-11-22T09:38:34","guid":{"rendered":"https:\/\/tobykskgd.life\/?p=3544"},"modified":"2025-11-22T17:40:32","modified_gmt":"2025-11-22T09:40:32","slug":"120","status":"publish","type":"post","link":"https:\/\/tobykskgd.life\/index.php\/120\/","title":{"rendered":"CS50P Problem Set 3"},"content":{"rendered":"\n

\u8bfe\u7a0b\uff1aCS50\u2019s Introduction to Programming with Python<\/a><\/p>\n\n\n\n

CS50 Python 2025 \u8bfe\u7a0b\u7684\u95ee\u9898\u96c6\u4f5c\u4e1a\u63d0\u4ea4\u3002\u4e3b\u8981\u63d0\u4f9b\u4e00\u4e2a\u672c\u4eba\u7684\u89e3\u9898\u601d\u8def\uff0c\u4ec5\u4f9b\u53c2\u8003\u3002<\/p>\n\n\n\n

Fuel Gauge<\/h3>\n\n\n\n

Problem<\/h4>\n\n\n\n

Fuel gauges indicate, often with fractions, just how much fuel is in a tank. For instance 1\/4 indicates that a tank is 25% full, 1\/2 indicates that a tank is 50% full, and 3\/4 indicates that a tank is 75% full.
\u71c3\u6cb9\u8868\u901a\u5e38\u4ee5\u5206\u6570\u8868\u793a\u6cb9\u7bb1\u5185\u7684\u71c3\u6cb9\u91cf\u3002\u4f8b\u5982\uff0c1\/4\u8868\u793a\u6cb9\u7bb1\u5df2\u6ee125%\uff0c1\/2\u8868\u793a\u6cb9\u7bb1\u5df2\u6ee150%\uff0c3\/4\u8868\u793a\u6cb9\u7bb1\u6cb9\u91cf\u4e3a75%\u3002<\/p>\n\n\n\n

In a file called fuel.py<\/code>, implement a program that prompts the user for a fraction, formatted as X\/Y<\/code>, wherein each of X<\/code> and Y<\/code> is a positive integer, and then outputs, as a percentage rounded to the nearest integer, how much fuel is in the tank. If, though, 1% or less remains, output E<\/code> instead to indicate that the tank is essentially empty. And if 99% or more remains, output F<\/code> instead to indicate that the tank is essentially full.
\u5728\u4e00\u4e2a\u540d\u4e3a fuel.py<\/code> \u7684\u6587\u4ef6\u4e2d\uff0c\u5b9e\u73b0\u4e00\u4e2a\u7a0b\u5e8f\uff0c\u63d0\u793a\u7528\u6237\u8f93\u5165\u4e00\u4e2a\u5206\u6570\uff0c\u683c\u5f0f\u4e3a X\/Y<\/code>\uff0c\u5176\u4e2d X<\/code> \u548c Y<\/code> \u5404\u4e3a\u6b63\u6574\u6570\uff0c\u7136\u540e\u8f93\u51fa\u6cb9\u7bb1\u5185\u71c3\u6599\u7684\u767e\u5206\u6bd4\uff0c\u56db\u820d\u4e94\u5165\u81f3\u6700\u63a5\u8fd1\u7684\u6574\u6570\u3002\u5982\u679c\u5269\u4f59 1%\u6216\u66f4\u5c11\uff0c\u5219\u8f93\u51fa E<\/code>\uff0c\u8868\u793a\u6cb9\u7bb1\u57fa\u672c\u7a7a\u4e86\u3002\u5982\u679c\u5269\u4f59 99%\u6216\u66f4\u591a\uff0c\u5219\u8f93\u51fa F<\/code>\uff0c\u8868\u793a\u6cb9\u7bb1\u57fa\u672c\u6ee1\u3002<\/p>\n\n\n\n

If, though, X<\/code> or Y<\/code> is not an integer, X<\/code> is greater than Y<\/code>, or Y<\/code> is 0<\/code>, instead prompt the user again. (It is not necessary for Y<\/code> to be 4<\/code>.) Be sure to catch any exceptions like ValueError<\/code><\/a> or ZeroDivisionError<\/code><\/a>.
\u4f46\u5982\u679c X<\/code> \u6216 Y<\/code> \u4e0d\u662f\u6574\u6570\uff0cX<\/code> \u5927\u4e8e Y<\/code>\uff0c\u6216 Y<\/code> \u662f 0<\/code>\uff0c\u5219\u518d\u6b21\u63d0\u793a\u7528\u6237\u3002\uff08Y<\/code> \u4e0d\u5fc5\u662f 4<\/code>\u3002\uff09\u52a1\u5fc5\u6355\u6349\u5f02\u5e38\uff0c\u5982 ValueError<\/code><\/a> \u6216 ZeroDivisionError<\/code><\/a>\u3002<\/p>\n\n\n\n

Submit<\/h4>\n\n\n\n
def main():\n    while True:\n        try:\n            fraction = input(\"Fraction: \")\n            x_str, y_str = fraction.split(\"\/\")\n            x = int(x_str)\n            y = int(y_str)\n            if 0 <= x <= y:\n                fuel = round(x \/ y * 100)\n                if fuel <= 1:\n                    print(\"E\")\n                elif fuel >= 99:\n                    print(\"F\")\n                else:\n                    print(f\"{fuel}%\")\n                break\n        except (ValueError, ZeroDivisionError):\n            pass\n\nmain()\n<\/code><\/pre>\n\n\n\n
Tips<\/h4>\n\n\n\n
int()<\/code> \u9047\u5230\u5e26\u5c0f\u6570\u70b9\u7684\u5b57\u7b26\u4e32\u4f1a\u629b\u51fa ValueError<\/strong><\/p>\n\n\n\n
\u793a\u4f8b\uff1a<\/p>\n\n\n\n
int(\"1.5\")<\/code><\/pre>\n\n\n\n
\u7ed3\u679c\uff1a<\/p>\n\n\n\n
ValueError: invalid literal for int() with base 10: '1.5'<\/code><\/pre>\n\n\n\n
\u539f\u56e0\uff1a<\/p>\n\n\n\n