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

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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

用字符串描述上一个字符串，即从左往右，每个数连续出现的次数

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#include <iostream>
#include <string>

using namespace std;

string read(string source) {
    string result;
    int cnt = 1;
    for (int i = 1; i < source.size(); i++) {
        if (source[i] != source[i - 1]) {
            result += to_string(source[i - 1] - '0') + to_string(cnt);
            cnt = 0;
        }
        cnt++;
    }
    result += to_string(*source.rbegin() - '0') + to_string(cnt);
    return result;
}

int main() {
    int d = 0, n = 0;
    scanf("%d %d", &d, &n);
    string s = to_string(d);
    for (int i = 1; i < n; i++)
        s = read(s);
    printf("%s", s.c_str());
    return 0;
}