把一个正整数进行因质分解 - 编程论坛

#2

东海ECS2023-04-01 14:53

您有什么问题吗?

#3

ysr28572023-04-01 16:47

回复楼主 zxq0103

def fen_jie(num): #  递归法
    for i in range(2,num+1):
        if i == num:
            print(i,end="")
            return
        elif num % i == 0:
            print(i,"* ",end="")
            fen_jie(num // i)
            return

def fen_jie2(num):
    while num != 1: # 循环判断法
        for i in range(2,num + 1):
            if num % i == 0:
                print(i,end="")
                num //= i
                if num != 1:
                    print(" * ",end="")
                break

if __name__ == "__main__":
    int_a= 142857
    print(int_a," = ",end="")
    fen_jie(int_a)
    print()
    print(int_a," = ",end="")
    fen_jie2(int_a)

这个程序可以运行，但要分解的数需要再程序中改顶，上面的结果如下：
...
142857  = 3 * 3 * 3 * 11 * 13 * 37
142857  = 3 * 3 * 3 * 11 * 13 * 37>>>

#4

东海ECS2023-04-01 19:01

可以改为接收输入:

程序代码：

def fen_jie(num): #  递归法
    for i in range(2,num+1):
        if i == num:
            print(i,end="")
            return
        elif num % i == 0:
            print(i,"* ",end="")
            fen_jie(num // i)
            return

def fen_jie2(num):
    while num != 1: # 循环判断法
        for i in range(2,num + 1):
            if num % i == 0:
                print(i,end="")
                num //= i
                if num != 1:
                    print(" * ",end="")
                break

if __name__ == "__main__":
    while True:
        try:
            int_a= 90
        except:
            print('error')
            break
        print(int_a," = ",end="")
        fen_jie(int_a)
        print()
        print(int_a," = ",end="")
        fen_jie2(int_a)

#5

zxq01032023-04-01 22:54

回复 2楼东海ECS

没什么问题,就是我新手,自己写的代码,让大神们给看一下,有没有需要改的地方
谢谢上面几位的指点

[此贴子已经被作者于2023-4-1 22:57编辑过]