pta 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：
例：D, D1--（1个D）, D111--（1个D 1个1）, D113--（1个D 3个1）, D11231--（1个D 2个1 1个3）

给出D和N
输出第n个数


测试得出 1 40

的长度为16138；可以确定范围


代码如下：

思路清晰就好做很多，我刚开始一直在一些细节上出问题，导致输出的很奇怪；注意int转char时+'0';

 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
char a[100000],b[100000];

int main()
{
    int d,n;
    cin >> d >> n;
    a[0] = d + '0';
    int co = 0;
    char st;
    if(n == 1)
    {
        cout << a[0] << endl;
        return 0;
    }
    else
    for(int i = 1;i < n;i++)
    {
        st = d + '0';//第一个数一直不会改变
        int num = 0;//记录个数
        int len = strlen(a);
        for(int j = 0;j < len;j++)
        {
            if(a[j] == st)
            num++;
            else
            {
                b[co++] = st;
                st = a[j];//更新st
                b[co++] = num + '0';
                num = 1;//更新num
            }
        }
        b[co++] = st;//最后一个在循环里无法录入，所以单独记录
        b[co++] = num + '0';
        for(int i1 = 0;i1 < co;i1++)
        a[i1] = b[i1];
        co = 0;
    }
    cout << a << endl;
    return 0;
}