1 solutions

  • 0
    @ 2025-11-5 16:15:35

    C++ :

    #include<iostream>
    using namespace std;
    int main()
    {
        int s=0;
        for (int i=1;i<=100;i++) if (i%2!=0 && i%3!=0) s+=i;
        cout<<s;
        return 0;
    }
    

    Pascal :

    var s,i:integer;
    begin
      s:=0;
      for i:=1 to 100 do
      if (i mod 2>0) and (i mod 3>0) then s:=s+i;writeln(s);
    end.
    
    
    
    
    • 1

    1~100之间既不能被2 整除,又不能被3整除的所有数之和

    Information

    ID
    17620
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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