1140 Look-and-say Sequence (20 分)

本文介绍了一种基于描述的整数序列生成方法,即Look-and-say序列,并提供了一个C++实现示例。该序列从给定的单个数字开始,后续每一项都是对其前一项的描述。

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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

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
using namespace std;
int d, n;
int main() {
	cin >> d >> n;
	string str;
	str.push_back(d + '0');
	for (int i = 1; i < n; i++) {
		string tmp;
		int cnt = 0;
		for (int j = 0; j < str.size(); j++) {
			cnt++;
			if (str[j] != str[j + 1]) {
				tmp += str[j];
				tmp.push_back(cnt + '0');
				cnt = 0;
			}
		}
		str = tmp;
	}
	printf("%s\n", str.c_str());
	return 0;
}

 

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