PAT甲级 1140

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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

题意：根据给定的首位数字和进行的次数，写出最后的结果。具体规则如下,第n次的结果要根据第n-1次的结果来写，从前往后每两个字符的位置上，第一个代表具体意义上的字符，第二个代表字符出现的次数。

解题思想：每一次循环都对初始输入的字符串进行变化，记录连续字符出现的次数，将字符及其字数保存到字符串中，最后进行输出即可。

#include <iostream>

using namespace std;

int main()
{
    int n;
    string s;
    cin>>s>>n;
    int i;
    for(i=1; i<n; i++)
    {
        char c=s[0];
        string x;
        int sum=0;
        for(int j=0; j<s.size(); j++)
        {
            if(s[j]==c)
            {
                sum++;
            }
            else
            {
                x.append(1,c);
                x.append(1,sum+'0');
                sum=1;
                c=s[j];
            }
        }
        x.append(1,c);
        x.append(1,sum+'0');
        s=x;

    }
    cout<<s<<endl;
    return 0;
}