1140 Look-and-say Sequence（20 分）
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根据题意就是显示几个数，下面就用自然数描述上一次的数即可。
如：1、11、21...表示1、1个1、2个1
#include<cstdio>
#include<string>
#include<iostream>
using namespace std;
int main(){
string s = "";
int n;
cin>>s>>n;
for(int i = 1;i < n;i++){
string p = "";
int len = s.size();
char be_c = s[0];
int pos = 0;
for(int j = 1;j <= len;j++){
if(j == len){
p += be_c;
p += (j - pos + '0');
}else if(be_c != s[j]){
p += be_c;
p += (j - pos + '0');
pos = j;
be_c = s[j];
}
}
s = p;
}
cout<<s<<endl;
return 0;
}