PAT甲1140字符串

PAT甲1140字符串

题目链接
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：
给两个数字D和n，第⼀个序列是D，后⼀个序列描述前⼀个序列的所有数字以及这个数字
出现的次数，⽐如D出现了1次，那么第⼆个序列就是D1，对于第⼆个序列D1，第三个序列这样描述：
D出现1次，1出现1次，所以是D111……以此类推，输出第n个序列～

题解：
⽤string s接收所需变幻的数字，每次遍历s，从当前位置i开始，看后⾯有多少个与s[i]相同，设j
处开始不相同，那么临时字符串 t += s[i] + to_string(j – i);然后再将t赋值给s，cnt只要没达到n次就继续
加油循环下⼀次，最后输出s的值～

#include<iostream>
using namespace std;
int main()
{
	string s;
	int n,j;
	cin>>s>>n;
	for(int cnt=1;cnt<n;cnt++)
	{
		string t;
		for(int i=0;i<s.length();i=j)
		{
			for(j=i;j<s.length()&&s[j]==s[i];j++);
				t+=s[i]+to_string(j-i);
		}
		s=t;
	}
	cout<<s;
	return 0;
}