1140 Lookandsay Sequence (20 分)（字符串处理）

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 =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～

原文链接：https://blog.csdn.net/liuchuo/article/details/79618698

题解

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,j;
    string s;
    cin>>s;
    cin>>n;
    for(int cnt=1;cnt<n;cnt++){
        string t;
        for(int i=0;i<s.size();i=j){
            for(j=i;j<s.size()&&s[j]==s[i];j++);
            t+=s[i]+to_string(j-i);
        }
        s=t;
    }
    cout<<s;
    return 0;
}

易懂版题解（来自博友）

#include<iostream>
using namespace std;
 
int main()
{
    string s;
    int n;
    cin>>s>>n;
    for(int i=1;i<n;i++){
        string nextStr;
        int cur=0;
        while(cur<s.length()){
            int next=cur+1;
            while(next<s.length()&&s[cur]==s[next]){
                next++;
            }
            nextStr+=s[cur]+to_string(next-cur);
            cur=next;
        }
        s=nextStr;
    }
    cout<<s;
}