u Calculate e
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 19465 | Accepted: 11362 | Special Judge | ||
Description
A simple mathematical formula for e is
e=Σ0<=i<=n1/i!
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.
Input
No input
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 Input
no input
Sample Output
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333 ...
#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 2002000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
LL n,m,j,i,t,k;
printf("n e
");
printf("- -----------
");
DD sum=0;
printf("0 1
");
for(i=1;i<=9;i++)
{
printf("%d ",i);
DD ans=1;
for(j=1;j<=i;j++)
ans*=j;
sum+=1/ans;
printf("%.10g
",sum+1);
}
return 0;
}