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  • 1140 Look-and-say Sequence

    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 }
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/12702184.html
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