u Calculate e

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

Total Submission(s): 39017    Accepted Submission(s): 17700

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

 

 

Source

#include
      
       double mul(double x){    double sum=1;    for(int i=1;i<=x;i++)        sum*=i;    return sum;}int main(){    int j;    double i,e=0,k;    printf("n e\n");    printf("- -----------\n");    printf("0 1\n");    printf("1 2\n");    printf("2 2.5\n");    for(j=0;j<10;j++)    {        e+=1/mul(j);        if(j>2)    printf("%d %11.9f\n",j,e);    }    return 0;}