1 条题解

  • 0
    @ 2025-4-12 21:41:02

    C :

    #include<stdio.h>
    int main(){
    int N;
    while(scanf("%d",&N)!=EOF){
        double z=2.0,m=1.0,t;
        int i;
        double sum=0.0;
        for(i=1;i<=N;i++){
            sum=sum+z/m;
            t=z;
            z=z+1;
            m=t;
            }
        printf("%.2f\n",sum);
    }
    return 0;
    }
    
    

    C++ :

    #include <iostream>
    #include <iomanip>
    using namespace std;
    
    int main()
    {
    	int n;
    	double sum;
    	while (cin>>n)
    	{
    		sum = 0;		
    		for (int i=1; i<=n; i++)
    		{			
    			sum += 1 + (double)1/i;
    		}
    		cout <<setprecision(2) <<std::fixed<< sum<<endl;
    	}
    	return 0;
    }
    
    

    Java :

    public class Main 
        {
                public static void main(String[] args)
                {
                  java.util.Scanner input = new java.util.Scanner(System.in);
                  
                  do {
                    int n = input.nextInt();
                    double sum = 0;
                    int n1 = 2;
                    int n2 = 1;
                    for (int i = 0;i < n;i++) {
                      sum += n1*1.0/n2;
                      
                      n1++; 
                      n2++;
                    }
                    long b = Math.round(sum*100);
                      double c = b/100.0;
                    
                    System.out.printf("%4.2f",sum);
                    System.out.print("\n");
                  }
                  while (input.hasNextInt()==true);
                }
    }
                      
                    
                    
    
    • 1

    信息

    ID
    1027
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者