PAT 甲级 1140 Look-and-say Sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/a845717607/article/details/88014688

本文探讨了看与说数列的生成原理，通过实例详细解释了数列的递推规则，并提供了一段AC代码实现，用于计算指定次数迭代后的数列结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1140 Look-and-say Sequence （20 point(s)）

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

经验总结：

这一题，由于不知道40次之后会有多大，所以使用string来进行操作比较稳妥，难度不大，就不多说啦~

AC代码

#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
int n;

int main()
{
	string str,temp;
	cin>>str>>n;
	for(int i=0;i<n-1;++i)
	{
		temp.clear();
		char str1[20];
		int num=1;
		for(int j=1;j<str.size();++j)
		{
			if(str[j]==str[j-1])
				++num;
			else
			{
				temp+=str[j-1];
				sprintf(str1,"%d",num);
				temp+=str1;
				num=1;
			}
		}
		temp+=str[str.size()-1];
		sprintf(str1,"%d",num);
		temp+=str1;
		str=temp;
	}
	printf("%s",str.c_str());
	return 0;
}