u Calculate e Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36073 Accepted Submission(s): 16277

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

#include <cmath>
#include <cstdio>
int main(){
printf("n e\n");
printf("- -----------\n");
printf("0 1\n");
double sum=1;
int product=1;
for(int i=1;i<=9;i++){
product*=i;
sum+=1/(double)product;
if(i>=3)
printf("%d %.9f\n",i,sum);
else if(i==2)
printf("%d %.1f\n",i,sum);
else if(i==1)
printf("%d %.0f\n",i,sum);
}
return 0;
}