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  • PAT 甲级 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

    题目大意:第n+1个串是第n个串的描述, 若第n个串是11122234,那么第n+1个串就是13233141,即1有3个、2有3个、3有1个、4有1个

    样例解释:暂无

    AC代码

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 
     4 int main()
     5 {
     6     int n; string s; cin >> s >> n;
     7     for(int i = 2; i <= n; i++)
     8     {
     9         string k; int top = 1;
    10         for(int j = 1; j <= s.length(); j++)
    11         {
    12             if(s[j] == s[j-1]) top++;
    13             else{
    14                 k += s[j-1] + to_string(top);
    15                 top = 1;
    16             }
    17         }
    18         s = k;
    19     }
    20     cout << s;
    21     return 0;
    22 }
    View Code


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  • 原文地址:https://www.cnblogs.com/kamisamalz/p/13612512.html
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