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

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探讨了Look-and-say序列的概念及其生成规则，通过输入特定的数字D和正整数N，计算并输出该序列的第N个元素。示例中详细解释了序列的形成过程，并提供了一个C++代码实现。

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

模拟题。思路比较简单。

注意点：

1. int2string()的实现；

2. 在nextString()遍历完成后，还需要针对字符串末尾的结果处理。

#include<iostream>
#include<cstring>
using namespace std;
string int2string(int n){
	string ans ="";
	do{
		ans = char(n%10+'0')+ans;
		n/=10;
	}while(n);
	return ans;
}
string nextString(string s){
	int cnt=0;string ans = "";
	for(int i=0;i<s.length();i++){
		if(i==0) cnt=1;
		else{
			if(s[i]==s[i-1]){
				cnt++;
			} 
			else{
				ans += s[i-1];
				ans += int2string(cnt);
				cnt=1;
			}
		}
	}
	ans+= s[s.length()-1];
	ans+= int2string(cnt);
	return ans;
}
int main(void){
	string s;int N;cin>>s>>N;
	if(N>=2){
		for(int i=0;i<N-1;i++){
			s = nextString(s);
		}
	} 
	cout<<s<<endl;
	return 0;
}