8 条题解

  • 1
    @ 2023-8-21 19:28:03

    #include <stdio.h> int main () { int n; int m=1; double x=0.0; scanf("%d",&n); for (int i=1;i<=n;i++) { m*=i; x+=1.0/m; } printf("%.2f",x); return 0; }

    • 1
      @ 2022-10-4 16:03:17
      #include<stdio.h>
      int main(){
      	
      	float i,term=0,sum=1;
      	int n;
      	scanf("%d",&n);
      	for(i=1;i<=n;i++){
      		sum*=i;
      	
      	term+=1/sum;
      }
      	printf("%.2f",term);
      	
      	
      	return 0;
      }
      
      • @ 2025-8-29 13:53:56

        牛逼

    • 0
      @ 2024-12-12 21:12:11

      #include <stdio.h> int main() { int n; double num=0; scanf("%d",&n);

      for(int i=1;i<=n;i++){ double x=1.0; for(int j=1;j<=i;j++){ x*=j; } num+=1/x; } printf("%.2f",num); return 0;

      }

      • 0
        @ 2024-4-24 22:13:23
        #include <stdio.h>
        int main()
        {
            int sp(int a);
            int a,b;
            scanf("%d",&a);
            double s=0.0;
            b=1;
            while(b<=a)
            {
                s=s+1.0/sp(b);
                b++;
            }
            printf("%.2lf",s);
            return 0;
        }
        int sp(int a)
        {
            int i=1,c=1;
            while(i<=a){c=c*i;i++;}
            return c;
        }//
        
        • 0
          @ 2023-10-18 11:52:32

          #include<stdio.h> int main() { int n; double sum=0.00,s=1; scanf("%d",&n); for(int i=1;i<=n;i++){ s*=i; sum+=1/s; } printf("%.2lf",sum); return 0; }

          • 0
            @ 2022-10-26 16:59:04

            #include<iostream> #include<iomanip> using namespace std;

            int main() { int n; cin>>n; int i=1; float b=1.00,sum=0.00; while(i<=n) { b=b*i; sum+=1/b; i++; } cout<<fixed<<setprecision(2)<<sum<<endl;

            return 0; }

            • 0
              @ 2022-10-6 10:48:07

              #include<stdio.h> int main(){ int n; double a=1.0,b=0.0; scanf("%d",&n); for(int i=1.0;i<=n;i++){ a=a*i ,b=b+1/a; } printf("%.2lf",b); return 0; }

              • 0
                @ 2022-10-3 23:40:47
                #include <stdio.h>
                int jiecheng(int k)
                {
                	if(k==1) return 1;
                	else return jiecheng(k-1)*k;
                }
                int main()
                {
                	int n; 
                	double sum=0;
                	scanf("%d",&n);
                	for(int k=1;k<=n;k++)
                		{
                		sum+= 1.0/jiecheng(k);
                		}
                	printf("%.2f",sum);	
                	return 0;
                }
                相比只加了一个阶乘(业余玩家写写玩)
                
                • 1

                信息

                ID
                34
                时间
                1000ms
                内存
                128MiB
                难度
                6
                标签
                递交数
                4305
                已通过
                1399
                上传者