Codeforces Round #527 (Div. 3) C. Prefixes and Suffixes (思维,字符串)

• 题意:给你某个字符串的\(n-1\)个前缀和\(n-1\)个后缀,保证每个所给的前缀后缀长度从\([1,n-1]\)都有,问你所给的子串是前缀还是后缀.

• 题解:这题最关键的是那两个长度为\(n-1\)的子串,我们只要判断哪个是前缀就行了,然后再遍历一遍所给的子串,用长度为\(n-1\)的前缀子串来判断是子串是前缀还是后缀.

• 代码:

  int n;
  string s[N];
  bool vis[N];
  int cnt;
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
      cin>>n;
  
      string s1="";
      string s2="";
      string pre;
      for(int i=1;i<=2*n-2;++i){
          cin>>s[i];
          if(s1.size()<s[i].size()) s1=s[i];
          else if(s1.size()==s[i].size()) s2=s[i];
      }
  
      for(int i=1;i<=2*n-2;++i){
          if(s1.substr(0,s[i].size())==s[i]) cnt++;
      }
  
      if(cnt>=n-1 && s1.substr(1,s1.size())==s2.substr(0,s2.size()-1)) pre=s1;
      else pre=s2;
  
      for(int i=1;i<=2*n-2;++i){
          if(s[i]==pre.substr(0,s[i].size()) && !vis[s[i].size()]){
              printf("P");
              vis[s[i].size()]=true;
          }
          else printf("S");
      }
  
      return 0;
  }