[已解决]课后作业求阶梯数的问题

spflmm

steps = 7
i = 1
FIND = False

while i < 100:
    if (steps % 7 == 0) and (steps % 6 == 5) and (steps % 5 == 4) and(steps % 3 == 2) and (steps % 2 == 1) :
        FIND = True
        break
        
    i = i + 1

if FIND == True:
    print('阶梯数是：', steps)
else:
    print('在程序限定的范围内找不到答案！')

为什么把steps能被7整除写成(steps % 7 == 0) 并且写到 if 里面不行，要写成下面这样才行？
为什么把steps能被7整除写成(steps % 7 == 0) 并且写到 if 里面不行，要写成下面这样才行？

steps = 7
i = 1
FIND = False

while i < 100:
    if (steps % 2 == 1) and (steps % 3 == 2) and (steps % 5 == 4) and (steps % 6 == 5):
        FIND = True
        break
    else:
        steps = 7 * (i + 1)
    i = i + 1

if FIND == True:
    print('阶梯数是：', steps)
else:
    print('在程序限定的范围内找不到答案！')

最佳答案

月排行榜 / 总排行榜

chinajz

2023-3-8 19:17:35

理解完全是对的，只是7的倍数效率似乎更好些。

1. #steps = 7
   
2. i = 7
   
3. FIND = False
   
4. 
   
5. while i < 120:
   
6.     if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
   
7.         FIND = True
   
8.         break
   
9.         
   
10.     i = i + 7
    
11. 
    
12. if FIND == True:
    
13.     print('阶梯数是：',i)
    
14. else:
    
15.     print('在程序限定的范围内找不到答案！')

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跳转到最佳答案楼层

chinajz

修改一下，也是可以的：

1. steps = 7
   
2. i = 1
   
3. FIND = False
   
4. 
   
5. while i < 100:
   
6.     steps = i*7
   
7.     if (steps % 7 == 0) and (steps % 6 == 5) and (steps % 5 == 4) and(steps % 3 == 2) and (steps % 2 == 1) :
   
8.         FIND = True
   
9.         break
   
10.         
    
11.     i = i + 1
    
12. 
    
13. if FIND == True:
    
14.     print('阶梯数是：', steps)
    
15. else:
    
16.     print('在程序限定的范围内找不到答案！')

复制代码

或者：

1. #steps = 7
   
2. i = 1
   
3. FIND = False
   
4. 
   
5. while i < 120:
   
6.     if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
   
7.         FIND = True
   
8.         break
   
9.         
   
10.     i = i + 1
    
11. 
    
12. if FIND == True:
    
13.     print('阶梯数是：',i)
    
14. else:
    
15.     print('在程序限定的范围内找不到答案！')

复制代码

chinajz

理解完全是对的，只是7的倍数效率似乎更好些。

1. #steps = 7
   
2. i = 7
   
3. FIND = False
   
4. 
   
5. while i < 120:
   
6.     if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
   
7.         FIND = True
   
8.         break
   
9.         
   
10.     i = i + 7
    
11. 
    
12. if FIND == True:
    
13.     print('阶梯数是：',i)
    
14. else:
    
15.     print('在程序限定的范围内找不到答案！')

复制代码

shigure_takimi

1. def stepNumbers(a,b):
   
2.     stepNumbers = []
   
3.     if a > b:
   
4.         a, b = b, a
   
5.     k = a//7
   
6.     for i in range(k*7, b, 7):
   
7.         if i%2 == 1 and i%3 == 2 and i%4 == 3 and i%5 == 4 and i%6 == 5:
   
8.             stepNumbers.append(i)
   
9.     if stepNumbers == []:
   
10.         print(f'[{a},{b}]之间没有阶梯数。')
    
11.     else:
    
12.         print(f'[{a},{b}]之间的阶梯数有{stepNumbers}')

复制代码