PAT甲级——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

首先看看样例是怎么变成1123123111的

1. 1
2. 11
3. 12
4. 1121
5. 122111
6. 112213
7. 12221131
8. 1123123111

本题要点：

1. string 只能用 cin cout 输入输出，不能使用 scanf 与 printf
2. t = t +…与 t+=… 后者效率较高
3. to_string()将数字转为字符串

#include <iostream>
#include <string>

using namespace std;

int main() {
    int N, j;
    string s;
    cin >> s >> N;
    for(int cnt = 1;cnt < N; cnt++){
        string t;
        for(int i = 0; i < s.length(); i = j){
            for(j = i; j < s.length() && s[j] == s[i]; j++);
            t += s[i] + to_string(j - i); //如果使用t=t+...会超时
        }
        s = t;
    }
    cout << s;
    return 0;
}