[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。

#steps = 7
i = 7
FIND = False

while i < 120:
    if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
        FIND = True
        break
        
    i = i + 7

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

跳转到最佳答案楼层

chinajz

修改一下，也是可以的：

steps = 7
i = 1
FIND = False

while i < 100:
    steps = i*7
    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！')

或者：

#steps = 7
i = 1
FIND = False

while i < 120:
    if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
        FIND = True
        break
        
    i = i + 1

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

chinajz

理解完全是对的，只是7的倍数效率似乎更好些。

#steps = 7
i = 7
FIND = False

while i < 120:
    if (i % 7 == 0) and (i % 6 == 5) and (i % 5 == 4) and(i % 3 == 2) and (i % 2 == 1) :
        FIND = True
        break
        
    i = i + 7

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

shigure_takimi

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