1140. Look-and-say Sequence (20)

1140. Look-and-say Sequence (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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：第n+1个串是第n个串的描述。   比如第n个串是11122234（随便写的），那么第n+1个串就是13233141，这个意思就是在第n个串中  1有3个， 2有3个， 3有1个， 4有1个， 所以是13233141。 

#include
using namespace std;
char d;
int n,len;
string s,temp,ans;
int main(){
    cin>>d>>n;
    s=d;
    int cnt=1,num=0;
    char t;
    while(cnt