题解 | #密码截取#
密码截取
https://www.nowcoder.com/practice/3cd4621963e8454594f00199f4536bb1
#include <iostream> #include <vector> #include <algorithm> using namespace std; int findMax(string temp, char a) { if (temp.size() == 1) return 2; for (int i = 0; i != temp.size(); ++i) { if (temp[i] == a) { if (i == temp.size() - 1) return 2; int k = temp.size() - i - 1; if ( k % 2 == 0) { string re = temp.substr(i + 1 + k / 2, k / 2); reverse(re.begin(), re.end()); if (temp.substr(i + 1, k/2) == re) return k+2; } else { string re = temp.substr(i + 1 + (k + 1) / 2, (k - 1) / 2); reverse(re.begin(), re.end()); if (temp.substr(i + 1, (k - 1) / 2) == re) return k + 2; } } } return 0; } int main() { string str; getline(cin,str); string temp; vector<int> result(str.size()+1,0); result[1] = 1; temp.push_back(str[0]); for(int i = 1;i!=str.size();++i) { result[i+1] = max(result[i],findMax(temp,str[i])); temp.push_back(str[i]); } cout<<result[temp.size()]; } // 64 位输出请用 printf("%lld")
动态规划思路,加入第i个字符,两种可能,此时字符串最大值不需要第i个字符,如果要用第i个字符,那么一定需要找到里面头尾是第i个字符的最大回文数。