A1035 Password (20 minutes | string processing, annotated in detail, logic analysis)

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This link: https://blog.csdn.net/qq_24452475/article/details/100568681

EDITORIAL

• Ideas analysis
  • After adjusting an array of strings in the string change vector plane memory ⾥
  • The number of array elements which is 0 or select a different logic branches
  • Outputting corresponding results
• Simple title, 10 minutes a title

Test Case

• input:
  3
  Team000002 Rlsp0dfa
  Team000003 perfectpwd
  Team000001 R1spOdfa
  
  output:
  2
  Team000002 RLsp%dfa
  Team000001 R@spodfa
  
  input:
  1
  team110 abcdefg332
  
  output:
  There is 1 account and no account is modified
  
  input:
  2
  team110 abcdefg222
  team220 abcdefg333
  
  output:
  There are 2 accounts and no account is modified
  

ac Code

• #include <iostream>
  #include <vector>
  using namespace std;
  
  int main()
  {
      int n;
      scanf("%d", &n);
      vector<string> v;
      for(int i=0; i<n; i++)
      {
          string name, s;
          cin >> name >> s;
          int len = s.length(), flag = 0;
          for(int j=0; j<len; j++)
          {
              switch(s[j])
              {
              case '1' :
                  s[j] = '@';flag = 1;break;
              case '0' :
                  s[j] = '%';flag = 1;break;
              case 'l' :
                  s[j] = 'L';flag = 1;break;
              case 'O' :
                  s[j] = 'o';flag = 1;break;
              }
          }
          if(flag)
          {
              string tmp = name + " " + s;
              v.push_back(tmp);
          }
      }
      int cnt = v.size();
      if(cnt != 0)
      {
          printf("%d\n", cnt);
          for(int i=0; i<cnt; i++)
              cout << v[i] << endl;
      }
      else if(n==1)
          printf("There is 1 account and no account is modified");
      else
          printf("There are %d accounts and no account is modified", n);
  
      return 0;
  }
  

Knowledge Point Summary

• String find and replace
  • s.replace(pos, n, s1) 用s1替换s中从pos开始（包括0）的n个字符的子串

  string ss = "Rlsp0dfa";
  if(ss.find('0'))
      cout << ss.replace(4, 1, "%") << endl;
  

  s.find(s1)         查找s中第一次出现s1的位置，并返回（包括0）
  s.rfind(s1)        查找s中最后次出现s1的位置，并返回（包括0）