PAT 1140. Look-and-say Sequence (20) – 甲级

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:给两个数字D和n,第一个序列是D,后一个序列描述前一个序列的所有数字以及这个数字出现的次数,比如D出现了1次,那么第二个序列就是D1,对于第二个序列D1,第三个序列这样描述:D出现1次,1出现1次,所以是D111……以此类推,输出第n个序列~
分析:用string s接收所需变幻的数字,每次遍历s,从当前位置i开始,看后面有多少个与s[i]相同,设j处开始不相同,那么临时字符串 t += s[i] + to_string(j – i);然后再将t赋值给s,cnt只要没达到n次就继续加油循环下一次,最后输出s的值~