1040 Longest Symmetric String (25分)

• dp动态规划

  1. \(dp[i][j]\)表示\(s[i] \to s[j]\)是否对称，是为\(1\)，否为\(0\)

  2. 状态转移方程：\(dp[i][j] = dp[i+1][j-1] \&\& (s[i]==s[j])\)

  3. 边界条件：\(dp[i][i]=1，dp[i][i+1]=(s[i]==s[i+1])\)

• 读取字符串时，使用cin.getline(char* s, streamsize n)代替gets(char* str)

  gets(char* str) is unsafe. It's recommended to use fgets(char* str, int num, FILE* stream) instead.^[1]

  However, I think it is trivial to process the string after using fgets().

  const int N = 200;
  char str[N];
  fgets(str, N, stdin);
  int i = 0;
  while(str[i] != '\n') 
        i++;
  str[i] = '\0'
  

  Thus, I use cin.getline() for its concision.

👉 code

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1. gets函数在pat里怎么用不了？ ↩︎