1 条题解

  • 0
    @ 2024-9-3 22:03:13

    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+m;
            m=t;
            }
        printf("%.2f\n",sum);
    }
    return 0;
    }
    
    

    C++ :

    #include <iostream>
    #include <iomanip>
    using namespace std;
    
    int main()
    {
    	int n;
    	double sum;
    	double F1, F2, F;
    	while (cin>>n)
    	{
    		F1 = 1;
    		F2 = 2;
    		sum = F2/F1;
    		
    		for (int i=2; i<=n; i++)
    		{
    			F = F1;
    			F1 = F2;
    			F2 = F1 + F;
    			sum += F2/F1;
    		}
    		cout <<setprecision(2) <<std::fixed<< sum<<endl;
    	}
    	return 0;
    }
    
    

    Java :

    import java.util.Scanner;
    public class Main{
      public static void main(String[] args){
      	Scanner input = new Scanner(System.in);
        while(input.hasNextInt()){
        	int N = input.nextInt();
          	int x,a = 1,b = 2;
          double sum = 0;
          for(x = 1;x <= N; x++){
    		sum = sum + ((double)b / a);
            int y = a;
              a = b;
            b = b + y;
          }
          String he = String.format("%.2f",sum);
          System.out.println(he);
        }
      }
    }
    
    • 1

    信息

    ID
    31
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者