思路很简单粗暴,判断要拨的号码在哪个数字上,接着把时间加入列表里,最后求和输出即可。 (当然,用“ord”函数也可以)

废话不多说,上代码(用的是Python3)

a=[[],
  ['A','B','C'],
  ['D','E','F'],
  ['G','H','I'],
  ['J','K','L'],
  ['M','N','O'],
  ['P','Q','R','S'],
  ['T','U','V'],
  ['W','X','Y','Z']]
b=input()
n=[]
for i in b:
    for j in range(1,len(a)):
        if i in a[j]:
            n.append(j+2)
print(sum(n))