• 1140 Look-and-say Sequence (20 分)


    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
    
     
    就是套循环了,n+1是用来描述n的。
     
     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 int n,m;
     4 int main(){
     5     cin >> n >> m;
     6     string s, st;
     7     s += n+'0';
     8     m--;
     9     while(m--){
    10         char c = s[0];
    11         int num = 1;
    12         for(int i = 1; i < s.length(); i++){
    13             if(c == s[i]){
    14                 num++;
    15             }else{
    16                 st += c;
    17                 st += '0'+num;
    18                 c = s[i];
    19                 num = 1;
    20             }
    21         }
    22         st += c;
    23         st += '0'+num;
    24         s = st;
    25         st = "";
    26     }
    27     cout << s <<endl;
    28     return 0;
    29 }
     
     
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  • 原文地址:https://www.cnblogs.com/zllwxm123/p/11286245.html
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