4 solutions

  • 4
    @ 2025-6-8 10:19:06
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int T;
    	cin>>T;
    	while(T--)
    	{
    		int a;
    		cin>>a;
    		int cnt=0;
    		for(int x=1;x<=a/x;x++) //枚举第一个数 
    		{
    			for(int y=1;y<=a/y;y++) //枚举第二个数 
    			{
    				if(x*x+y*y==a)
    				{
    					cnt++;
    				}
    			}
    		}
    		if(cnt>0) cout<<"Yes"<<endl;
    		else cout<<"No"<<endl;
    	}
    	return 0;
    }
    
    
    • 1
      @ 2026-2-26 11:11:01
      #include<bits/stdc++.h>
      using namespace std;
      int main()
      {
          int n;
          cin>>n;
          for(int i=1;i<=n;i++)
          {
              int a;
              cin>>a;
              int s=0;
              for(int x=1;x*x<=a;x++)
              {
                  for(int y=1;y*y<=a;y++)
                  {
                      if(x*x+y*y==a)
                      {
                          s=1;
                      }
                     
                  }
              }
              if(s==1)
              {
                  cout<<"Yes"<<endl;
              }
              else
              {
                  cout<<"No"<<endl;
              }
          }
          return 0;
      } 
      
      • -1
        @ 2025-11-15 15:59:04

        压力

        • -1
          @ 2025-6-8 10:19:37
          #include<bits/stdc++.h>
          using namespace std;
          int a[1111111];
          int main()
          {
              int T;cin>>T;
              while(T--)
              {
                  int n,cnt=0;cin>>n;
                  for(int i=1;i<=n/i;i++)
                  {
                      for(int j=1;j<=n/j;j++)
                      {
                          if(i*i+j*j==n)
                          {
                              cnt++;
                          }
                      }
                  }
                  if(cnt==0) cout<<"No"<<endl;
                  else cout<<"Yes"<<endl;
              }
              return 0;
          }
          
          • 1

          Information

          ID
          2179
          Time
          1000ms
          Memory
          256MiB
          Difficulty
          1
          Tags
          # Submissions
          112
          Accepted
          30
          Uploaded By