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
Sample Output:
1123123111题目大意
给定D和N，要求输出N次之后对D的描述。
描述方法为：
1
11 上一个数包含1，且只有1个
12 上一个数包含1，且有2个
1121 上一个数包含1，有1个；包含2，有1个
122111 上一个数包含1，有2个；包含2，有1个；包含1，有1个
112213 上一个数包含1，有1个；包含2，有2个；包含1，有3个
。。。。。。。。。。。。。。。。。。。
分析
使用递归的describe函数进行描述，当进入最后一次描述时，返回结果。
每次进行描述的时候，从0开始遍历整个字符串，遇到第一个不相同的字符时，完成前面的字符描述。要注意遇上最后一个字符的时候，要进行描述并结束。
Python实现
def describe(d, k):
if k == 1:
return d
else:
temp, start = '', 0
for x in range(len(d)):
if d[x] != d[start]:
temp = temp + d[start] + str(x - start)
start = x
if x == len(d) -1:
temp = temp + d[start] + str(x - start + 1)
return describe(temp, k - 1)
if __name__ == "__main__":
line = input().split(" ")
result = describe(line[0], int(line[1]))
print(result)