3 solutions

  • 1
    @ 2022-4-5 1:32:47

    还好

    #include <iostream>
    using namespace std;
    int main()
    {
        int T = 0;
        cin >> T;
        while(T--)
        {
            int n =0;
            cin >> n;
            if(n<60)    cout << "0.0" << endl;
            else
            {
                double d = 1.0 * (n-60)/10+1;
                printf("%.1f", d);
                printf("\n");
    
            }
        }
        return 0;
    }
    
    • 0
      @ 2022-11-26 18:16:54
      #include<iostream>
      #include<iomanip>  // --->setprecision所需头文件
      using namespace std;
      
      int main(){
      	int T, n;
      	double ans;
      	cin >> T;
      	for(int i = 1; i <= T; i++){
      		cin >> n;
      		
      		if(n < 60)
      			cout << "0.0" << endl;
      		else{
      			ans = 1.0 * (n - 60) / 10 + 1;
      			cout << setprecision(1) << fixed<< ans << endl;	//保留一位小数
      		}		
      	}
      	
      	return 0;
      }
      
      • 0
        @ 2022-3-7 22:11:56
        #include <stdio.h>
        #include <algorithm>
        #include <math.h>
        using namespace std;
        int n;
        int main() {
            scanf("%d",&n);
            while(n--)
            {
                double tmp;
                scanf("%lf",&tmp);
                if(tmp<60)
                    printf("0.0\n");
                else
                {
                    double ans=(tmp-60)/10+1;
                    printf("%.1lf\n",ans);
                }
        
            }
        
        
            return 0;
        }
        
        
        • 1

        Information

        ID
        23
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        7
        Tags
        # Submissions
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        Accepted
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