u Calculate e

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 32719 Accepted Submission(s): 14683

Problem Description

A simple mathematical formula for e is



where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output

   
   
   
   

    
    
    
    n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    源代码一：
   
   
   
   
   
   
   
   

    
    
    
    #include <stdio.h> #include <stdlib.h> double fac(int n) {    double c; if(n==1 || n==0) {       c=1;  return(c); } else {       c=fac(n-1)*n;
   
   
   
   
   
   
   
   

    
    
    
     return(c); } } int main() {       double sum,t; int i,n,j=3; printf("n e\n");    printf("- -----------\n");    printf("0 1\n");    printf("1 2\n");    printf("2 2.5\n");    while(j<10)    {      for(i=0,sum=0;i<=j;i++)       {     t=fac(i);     sum=sum+1/t;      }       printf("%d %11.9f\n",j,sum);        j++;                                     }    system("pause");  return 0; } 
   
   
   
   
   
   
   
   

    
    
    
    源代码二：
   
   
   
   
   
   
   
   

    
    
    
    #include <stdio.h> #include <stdlib.h> int main() {   double arr[10] = {1},result;   int i = 1,j = 3;   while(i < 10)   {     arr[i]=i*arr[i-1];     i++;   }   printf("n e\n");   printf("- -----------\n");   printf("0 1\n");   printf("1 2\n");   printf("2 2.5\n");   result = 2.5;   while(j < 10)   {     result = result + 1/arr[j];     printf("%d %11.9f\n",j,result);     j++;   }   system("pause");   return 0; }