7 solutions

  • 3
    @ 2024-3-15 14:59:01
    枚举奇偶,向两边扩张
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	string s;
    	getline(cin,s);
    	int res=0;
    	for(int i=0;i<s.size();i++)
    	{
    		int l=i-1,r=i+1; //回文串长度是奇数的情况 
    		while(l>=0&&r<s.size()&&s[l]==s[r]) l--,r++; 
    		res=max(res,r-l-1);
    		l=i,r=i+1;       //回文串长度是偶数的情况 
    		while(l>=0&&r<s.size()&&s[l]==s[r]) l--,r++;
    		res=max(res,r-l-1);
    	}
    	cout<<res;
    	return 0;
    }
    
    • 1
      @ 2025-7-23 14:42:51

      • 1
        @ 2025-3-30 15:18:50
        #include<bits/stdc++.h>
        using namespace std;
        int main()
        {
        	string s;
        	getline(cin,s);
        	int res=0;
        	for(int i=0;i<s.size();i++)
        	{
        		int l=i-1,r=i+1; 
        		while(l>=0&&r<s.size()&&s[l]==s[r]) l--,r++; 
        		res=max(res,r-l-1);
        		l=i,r=i+1;       
        		while(l>=0&&r<s.size()&&s[l]==s[r]) l--,r++;
        		res=max(res,r-l-1);
        	}
        	cout<<res;
        	return 0;
        }
        
        
        • 0
          @ 2025-7-23 14:47:41

          • 0
            @ 2025-7-23 14:47:20

            • -1
              @ 2025-7-23 14:46:55

              • -3
                @ 2024-9-18 20:21:07
                #include<bits/stdc++.h>
                using namespace std;
                int main(){
                	string shuru;
                	cin>>shuru;
                	int jishu=0;
                	for(int i=0;i<shuru.size();i++)
                	{
                		int qian=i-1,hou=i+1;
                		while(shuru[qian]==shuru[hou]&&qian>-1&&hou<shuru.size())
                		{
                			qian--;
                			hou++;
                		}
                		jishu=max(hou-qian-1,jishu);
                		
                		qian=i;
                		hou=i+1;
                		while(shuru[qian]==shuru[hou]&&qian>-1&&hou<shuru.size())
                		{
                			qian--;
                			hou++;
                		}
                		jishu=max(hou-qian-1,jishu);
                	}
                	cout<<jishu;
                	return 0;
                }
                
                
                
                • 1

                Information

                ID
                79
                Time
                1000ms
                Memory
                256MiB
                Difficulty
                3
                Tags
                # Submissions
                88
                Accepted
                22
                Uploaded By