首页检查以下代码问题并修改，def max_assignments(A): A = sorted(A, key=lambda x: x[1]) current_day = 1 count = 0 for duration, deadline in A: if current_day + duration - 1 <= deadline: count += 1 current_day += duration return count A = [[2, 4], [3, 5], [1, 2], [4, 7], [1, 1]]

检查以下代码问题并修改，def max_assignments(A): A = sorted(A, key=lambda x: x[1]) current_day = 1 count = 0 for duration, deadline in A: if current_day + duration - 1 <= deadline: count += 1 current_day += duration return count A = [[2, 4], [3, 5], [1, 2], [4, 7], [1, 1]]

时间: 2023-04-09 08:03:23 浏览: 81

找质数代码matlab-Number_Theory_and_cryptography_assignments:Number_Theory_an

找质数代码matlab NTC任务1 贡献者： Govind Jeevan 16CO221 Palak Singhal 16CO129 问题陈述：编写MATLAB代码，以正确的过程演示费马小定理a ^（p-1）≡1（modp）。费马小定理： a^(p-1) ≡ 1 mod p 其中a不能被p整除，而p是质数。我们提出了四种方法来证明费马小定理示范一：结论也可以写成 remainder(a^(p-1), p) = 1, 表示将a ^（p-1）^除以p的余数是1，或a ^（p-1）^ 1 mod p。其中a不能被p整除。输入：质数的格式。（例如4x + 3）被测x范围的上限和下限算法： 1. p = 4x+1 2. a^(p-1) is calculated 3. Remainder on division by p. 4. Verify if the remainder is 1 . 5. If remainder != 1, it is not a prime 6. If remainder ==1, it is a prime or a pseudoprime

修改后的代码如下： def max_assignments(A): A = sorted(A, key=lambda x: x[1]) current_day = 1 count = 0 for duration, deadline in A: if current_day + duration - 1 <= deadline: count += 1 current_day += duration return count A = [[2, 4], [3, 5], [1, 2], [4, 7], [1, 1]] print(max_assignments(A)) 输出结果为3，表示最多可以完成3项任务。

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