A1140 Look-and-say Sequence (20 point(s))

递推字符串

1. 原文

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

2. 解析

字符串string – STL

比较字符串记录重复值cnt++,当字符不相等时记录当前字符和cnt

if (str[j]==str[j+1]){
    cnt++;
}else{
    ans+=str[j];
    ans+=cnt+'0';
    cnt=1;
}

3. AC代码

#include<iostream>
#include<string>
using namespace std;
int main()
{
	int d,n;
	scanf("%d%d",&d,&n);
	string str;
    if (n==1)
    {
    	printf("%d",d);
    	return 0;
    }
	  str=d+'0';
    for (int i = 0; i < n-1; ++i)
    {
    	int cnt=1; string ans;
    	for (int j=0;j<str.length();j++)
    	{
    		if (str[j]==str[j+1])
    		{
    			cnt++;
    		}else{
    			ans+=str[j];
    			ans+=cnt+'0';
    			cnt=1;
    		}
    	}
    	str=ans;
    }
    printf("%s\n",str.c_str());

	return 0;
}