3 solutions

  • 0
    @ 2025-7-22 10:31:32
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int n,m;
    	cin>>n>>m;
    	int sum=0;
    	for(int i=1;i<=n;i++)
    	{
    		int x;
    		cin>>x;
    		if(i%2==0) sum--;
    		sum+=x;
    	}
    	if(sum<=m)
    	{
    		cout<<"Yes";
    	}
    	else
    	{
    		cout<<"No";
    	}
    	return 0;
    }
    
    • -2
      @ 2025-6-23 19:56:26
      #include<bits/stdc++.h>
      using namespace std;
      int a[1111];
      int main()
      {
          int n,m;
          cin>>n>>m;
          for(int i=1;i<=n;i++)
          {
              cin>>a[i];
          }
          int sum=0;
          for(int i=1;i<=n;i++)
          {
              sum+=a[i];
              if(i%2==0&&i>=2)
              {
                  sum--;
              }
          }
          if(sum>m)
          {
              cout<<"No";
              return 0;
          }
          else if(sum<=m) cout<<"Yes";
          return 0;
      }
      

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      • -3
        @ 2025-5-17 9:20:29
        using namespace std;
        const int N=10010;
        int a[N];
        int main()
        {
        	int n,x;
        	cin>>n>>x;
        	for(int i=1;i<=n;i++)
        	{
        		cin>>a[i];
        	}
        	int q=0;
        	for(int j=1;j<=n;j++)
        	{
        		if(j%2!=0)
        		{
        			q=q+a[j];
        		}
        		else
        		{
        			q=q+a[j]-1;
        		}
        	}
        	if(q<=x)
        	{
        		cout<<"Yes";
        	}
        	else
        	{
        		cout<<"No";
        	}
        	return 0;
        }
        
        • 1

        [ABC209B] 你能全部买下吗?(Can you buy them all?)

        Information

        ID
        2335
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        1
        Tags
        (None)
        # Submissions
        97
        Accepted
        31
        Uploaded By