3 solutions

  • 0
    @ 2025-9-17 18:14:54
    
    

    #include<bits/stdc++.h> using namespace std; int main() { int n; string a; cin>>n>>a; for(int i=0;i<a.size();i++){ a[i]='a'+(a[i]-'a'+n)%26; } cout<<a; return 0; }

    • 0
      @ 2025-8-29 16:29:51

      • 0
        @ 2025-8-28 16:45:32
        
        #include<bits/stdc++.h>
        using namespace std;
        int main() {
        int n;
        string m;
        cin>>n>>m;
        for(int i=0;i<m.size();i++){
            m[i]='a'+(m[i]-'a'+n)%26;
        }
        cout<<m;
            return 0;
        }
        
        • 1

        【深基6.例2】小书童——凯撒密码(caesar)

        Information

        ID
        101
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        7
        Tags
        # Submissions
        76
        Accepted
        15
        Uploaded By