1140 Look-and-say Sequence

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本文介绍了一种特殊的整数序列——Look-and-say数列，并提供了一个C++实现方案来生成给定数字D的第N项。该数列通过描述前一项的构成来生成下一项。

分析：简单题，一次过，注意maxn为40，所以用string存储比较好。
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
代码

#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;

const int maxn = 41;
int main(){
    int D, N;
    scanf("%d %d", &D, &N);
    string strD = to_string(D);
    string bfstr = strD;
    for(int i = 0; i < N - 1; i++){
    	int count = 1;
    	string newstr = "";
        for(int j = 0; j < bfstr.length(); j++){
        	if(j + 1 < bfstr.length()){
        		if(bfstr[j] == bfstr[j + 1]){
        			count++;
				}else{
					newstr += bfstr[j] + to_string(count);
					count = 1;
				}
			}else{
				newstr += bfstr[j] + to_string(count);
				count = 1;
			}
		}
        bfstr = newstr;
    }
    cout << bfstr;
}