2 solutions

  • 1
    @ 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
      72
      Accepted
      24
      Uploaded By