题目
Look-and-say sequence is a sequence of integers as the following:
D, D1, D111, D113, D11231, D112213111, ... where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is one D in the 1st number, and hence it is D1; the 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to 11), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, two 1's, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now you are supposed to calculate the Nth number in a look-and-say sequence of a given digit D.
Input Specification: Each input file contains one test case, which gives D (in [0, 9]) and a positive integer N (≤ 40), separated by a space.
Output Specification: Print in a line the Nth number in a look-and-say sequence of D.
Sample Input:
1 8
结尾无空行
Sample Output:
1123123111
结尾无空行
解题思路
startInt, round = map(str, input().split())
# startInt, round = map(str, "1 8".split())
res = startInt
# res = str(1231)
for _ in range(int(round)-1):
# resList = list(res)
resList = res
tempStr = None
tempCount = 0
res = ""
for val in resList:
# for index,val in enumerate(resList):
if val == tempStr:
tempCount += 1
else:#如果不相等，就把上一个输出并新赋值
if tempStr != None:
res += tempStr + str(tempCount)
tempStr = val
tempCount = 1
#处理最后一个符号
res += tempStr + str(tempCount)
print(res)
