6 条题解

  • 0
    @ 2023-9-18 21:37:53
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        double c , a = 2.50;
        cin >> c;
        if(c <= 10)
        {
            printf("%.2f",a);
        }
        else
        {
            c = c - 10;
            c = c * 1.50;
            a = c + a;
            printf("%.2f",a);
        }
        return 0;
    }
    

    A

    • -1
      @ 2023-7-7 12:51:21
      #include<iostream>
      #include<iomanip>
      using namespace std;
      int main()
      {
      	int n;
      	cin>>n;
      	if(n<=10)
      	{
      		cout<<fixed<<setprecision(2)<<2.5;
      	}
      	else
      	{
      		cout<<fixed<<setprecision(2)<<(n-10)*1.5+2.5;
      	}
      	return 0;
      }
      
      • -1
        @ 2022-12-23 13:51:27
        #include <iostream>
        #include <iomanip>
        using namespace std;
        int main()
        {
        	float x,n;
        	cin >> n;
        	if (n <= 10)
            {
        		x = 2.5;
        	}
            if (n > 10)
            {
        		x = 2.5 + (n - 10) * 1.5;
        	}
        	cout << fixed << setprecision(2) << x << endl;
            return 0;
        }
        
        • -2
          @ 2023-8-31 15:15:05

          //什么是赞?哪位大佬可以点给我看一下?求求了❤️

          #include <bits/stdc++.h>
          using namespace std;
          int main(){
              int n;
              cin >> n;
              if(n <= 10)
              {
                  cout << fixed << setprecision(2) << 2.5;
              }
              if(n > 10)
              {
                  cout << fixed << setprecision(2) << (n - 10) * 1.5 + 2.5;
              }
              return 0;
          
          • -2
            @ 2023-4-2 18:51:17
            #include<bits/stdc++.h>
            using namespace std;
            int main()
            {
                int x;
                cin>>x;
                if (x<=10)
                    cout<<fixed<<setprecision(2)<<2.5;
                else
                    cout<<fixed<<setprecision(2)<<1.5*(x-10)+2.5;
                return 0;
            }
            
            • -2
              @ 2022-11-10 13:21:56

              号没了

              #include <bits/stdc++.h>
              using namespace std;
              int main()
              {
                  double x;
                  cin >> x;
                  if(x<=10)
                  {
                      cout << fixed << setprecision(2) << 2.5;
                  }
                  else
                  {
                      cout << fixed << setprecision(2) << 2.5+(x-10)*1.5;
                  }
              }
              
              • 1

              信息

              ID
              38
              时间
              1000ms
              内存
              16MiB
              难度
              4
              标签
              递交数
              320
              已通过
              147
              上传者