1140 Look-and-say Sequence

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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

题意：

　　用后一个数描述前一个数，例如：1，11, 12, 1121，第一个数是给定的，第二个数是说：第一个数由1个1组成；第三个数是说：第二个数由2个1组成；第四个数是说：第三个数由1个1和1个2组成。

思路：

　　首先将数字转成字符串，末尾加一个空格（方便输出），ans用来存储这次生成的字符串。注意有组测试是测的第一个数字，直接输出就行了。

Code：

 1 #include <iostream>
 2 #include <string>
 3 
 4 using namespace std;
 5 
 6 int main() {
 7     int D, N, count;
 8     cin >> D >> N;
 9     if (N == 1) {
10         cout << D << endl;
11         return 0;
12     }
13     string ans = to_string(D) + "1";
14     string str = ans + " ";
15     for (int i = 3; i <= N; ++i) {
16         count = 1;
17         ans = "";
18         for (int j = 1; j < str.length(); ++j) {
19             if (str[j] == str[j - 1])
20                 count++;
21             else {
22                 ans += str[j - 1] + to_string(count);
23                 count = 1;
24             }
25         }
26         str = ans + " ";
27     }
28     cout << ans << endl;
29     return 0;
30 }

1140 Look-and-say Sequence

标签：space   win   ati   style   clipboard   number   nbsp   therefore   wing   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/12702184.html