5 solutions

  • 10
    @ 2024-5-15 14:00:17
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int n;
    	cin>>n;
    	int sum=0;
    	for(int i=1;i<=n;i++)
    	{
    		//1.第一种情况判断 
    		if(i%7==0) continue;//说明i是7的倍数
    		//2.第二种情况判断 
    		int cnt=0;
    		int t=i;
    		while(t) //循环拆数模板 
    		{
    			int ge=t%10;
    			if(ge==7) cnt++;
    			t/=10;	
    		} 
    		if(cnt!=0) continue;//说明10进制中某一位有7
    		//说明所有情况都满足 
    		
    		sum+=i*i; 
    	}
    	cout<<sum;
    	return 0;
    }
    
    • 2
      @ 2026-8-13 15:04:18
      #include<bits/stdc++.h>
      using namespace std;
      int main()
      {
      	int n;
      	cin>>n;
      	int s=0;
      	for(int i=1;i<=n;i++)
      	{
      		if(i%7==0) continue;
      		int cnt=0;
      		int t=i;
      		while(t)
      		{
      			int ge=t%10;
      			if(ge==7) cnt++;
      			t/=10;
      		}
      		if(cnt!=0) continue;
      		s+=i*i;
      	}
      	cout<<s;
      	return 0;
      }
      
      • -2
        @ 2026-4-19 16:33:52

        #include<bits/stdc++.h>

        using namespace std;

        int main(){

        int n,x=0;
        
        
        scanf("%d",&n);
        
        for(int i=1;i<=n;i++){
        
            if(i%7==0) continue;
            
            int j=i;
            
            while(j!=0){
            
                if(j%10==7) break;
        
                
                j/=10;
                
            }
            
            if(j==0) x+=pow(i,2);
            
        }
        
        printf("%d",x);
        

        return 0;

        }

        • -3
          @ 2024-11-4 18:45:59
          #include<bits/stdc++.h>
          using namespace std;
          int main()
          {
          	int n;
          	cin>>n;
          	int sum=0;
          	for(int i=1;i<=n;i++)
          	{
          		//1.第一种情况判断 
          		if(i%7==0) continue;//说明i是7的倍数
          		//2.第二种情况判断 
          		int cnt=0;
          		int t=i;
          		while(t) //循环拆数模板 
          		{
          			int ge=t%10;
          			if(ge==7) cnt++;
          			t/=10;	
          		} 
          		if(cnt!=0) continue;//说明10进制中某一位有7
          		//说明所有情况都满足 
          		
          		sum+=i*i; 
          	}
          	cout<<sum;
          	return 0;
          }
          
          • -3
            @ 2024-10-19 12:04:37
            #include<bits/stdc++.h>
            using namespace std;
            int main()
            {
                int n,s=0;
                bool flag;
                cin>>n;
                for(int i=1;i<=n;i++)
                {
                    if(i%7==0) continue;
                    int temp=i;
                    while(temp)
                    {
                        if(temp%10==7)
                        {
                            flag=true;
                            break;
                        }
                        else flag=false;
                        temp/=10;
                    }
                    if(flag) continue;
                    s+=i*i;
                }
                cout<<s;
                return 0;
            }
            
            • 1

            Information

            ID
            889
            Time
            1000ms
            Memory
            256MiB
            Difficulty
            1
            Tags
            (None)
            # Submissions
            228
            Accepted
            83
            Uploaded By