【发布时间】:2016-02-06 23:22:32
【问题描述】:
用户将输入 10 个整数,作为回报,程序将告诉用户输入的 10 个整数中最大的奇数是哪个。我在下面粘贴了我的工作脚本。我的工作的问题是,当最大的数字是偶数时,它会返回错误,我怎样才能让 python 忽略最大的偶数并继续下一个最大的奇数。
n1 = int(input('enter integer 1: '))
n2 = int(input('enter integer 2: '))
n3 = int(input('enter integer 3: '))
n4 = int(input('enter integer 4: '))
n5 = int(input('enter integer 5: '))
n6 = int(input('enter integer 6: '))
n7 = int(input('enter integer 7: '))
n8 = int(input('enter integer 8: '))
n9 = int(input('enter integer 9: '))
n10 = int(input('enter integer 10: '))
if n1%2!=0:
a=n1
if n2%2!=0:
b=n2
if n3%2!=0:
c=n3
if n4%2!=0:
d=n4
if n5%2!=0:
e=n5
if n6%2!=0:
f=n6
if n7%2!=0:
g=n7
if n8%2!=0:
h=n8
if n9%2!=0:
i=n9
if n10%2!=0:
j=n10
if n1>n2 and n1>n3 and n1>n4 and n1>n5 and n1>n6 and n1>n7 and n1>n8 and n1>n9 and n1>n10:
print(n1, 'is the largest odd number')
if n2>n1 and n2>n3 and n2>n4 and n2>n5 and n2>n6 and n2>n7 and n2>n8 and n2>n9 and n2>n10:
print(b, 'is the largest odd number')
if n3>n1 and n3>n2 and n3>n4 and n3>n5 and n3>n6 and n3>n7 and n3>n8 and n3>n9 and n3>n10:
print(c, 'is the largest odd number')
if n4>n1 and n4>n2 and n4>n3 and n4>n5 and n4>n6 and n4>n7 and n4>n8 and n4>n9 and n4>n10:
print(d, 'is the largest odd number')
if n5>n1 and n5>n2 and n5>n3 and n5>n4 and n5>n6 and n5>n7 and n5>n8 and n5>n9 and n5>n10:
print(e, 'is the largest odd number')
if n6>n1 and n6>n2 and n6>n3 and n6>n4 and n6>n5 and n6>n7 and n6>n8 and n6>n9 and n6>n10:
print(f, 'is the largest odd number')
if n7>n1 and n7>n2 and n7>n3 and n7>n4 and n7>n5 and n7>n6 and n7>n8 and n7>n9 and n7>n10:
print(g, 'is the largest odd number')
if n8>n1 and n8>n2 and n8>n3 and n8>n4 and n8>n5 and n8>n6 and n8>n7 and n8>n9 and n8>n10:
print(h, 'is the largest odd number')
if n9>n1 and n9>n2 and n9>n3 and n9>n4 and n9>n5 and n9>n6 and n9>n7 and n9>n8 and n9>n10:
print(i, 'is the largest odd number')
if n10>n1 and n10>n2 and n10>n3 and n10>n4 and n10>n5 and n10>n6 and n10>n7 and n10>n8 and n10>n9:
print(j, 'is the largest odd number')
【问题讨论】:
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返回错误 - 我们必须猜测报告的错误是什么,还是您会告诉我们?
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尝试阅读列表和循环。它们将使您的程序更简单