"C + +" Calculate The exact number of E.

As we know,the Natural Constant,which usually called E,has both main ways of expressing:

1. \ (\lim_{n \to \infty}\sum_{i=0}^{n} (\frac{1}{i!}) \)

2. \ (\lim_{n \to \infty} (1+\frac{1}{n}) ^{n} \)

Let's have a try,the following-a-Python (Version 3.4) program shows you the cruel fact:(

2.7181459268249255

	number	the level of deviation
number of Formula one	2.7182818284590455	\ ({10}^{-16} \)
number of Formula both	\ ({10}^{-4} \
the exact number of e	2.7182818284590452	------

So obviously,the convergence rate of Formula One are much more fast-speed. Also the calculate time it costs is not far more than Formule one.

So take action.:)

Another fact is,what most of us care most are how many accurate digits we can reach,not what big or small the \ (n \) is.

Though The high accuration calculation are required, the program still has both variables: \ (s \) and \ (t \), which is U Sually used to store the sum and the factorial. When the \ (n \) are large enough, the very digit of s can be stable, which means the same digit of T are \ (0 \), to be SAF Er, we should put another sereval \ (0 \) behind it, which won ' t cost much more time.

The One billion (\ ({10}^{9} \)) binary system should is used in order to minimize the time use.

So are the program:

#include <cstdio>#include<iostream>#include<algorithm>#include<string.h>#pragmaWarning (disable:4996)#definell Long Longusing namespacestd;Const intmaxn=100010; ll*a,*b;intN,m,x;ll p;voidDivintT) {//High-accuration Divisionll s=0;  for(inti=x;i<=n;i++) {s=s*p+B[i]; B[i]=s/T; S%=T; }     while(b[x]==0) x + +;}stringOP (ll x,BOOLfx=false){//Translate 10^9 number into string    strings="";  for(intI=1; i<=m;i++){        strings1=""; S1=(Char) (%Ten+ -); S1+=s;s=S1; X=x/Ten; }    if(FX) {intI=0;  while(s.at (i) = ='0') i++; S=s.substr (i); }    returns;}intMain () {CIN>> n;n++; M=9;intv=n;n=n/m+5; intnn=n;n=n+8;//Some more space should be sparedp=1; for(intI=1; i<=m;i++) p*=Ten; A=Newll[n+1];b=Newll[n+1];  for(intI=2; i<=n;i++) a[i]=0; a[1]=1; a[n+1]=1;  for(intI=2; i<=n;i++) b[i]=0; b[1]=1; b[n+1]=1;//Initialize A and B, which means s and T.    intt=1; x=1;  while(x<=nn) {div (t+ +); ll k=0;  for(inti=n;i>=1; i--) {A[i]+=b[i]+K; K=a[i]/p;a[i]%=p; }        inti=x-1;  while(k!=0) {A[i]+=k;k=a[i]/p; A[i]%=p;i--;//High-accuraton plus        }    }    strings1="";  for(intI=1; i<=nn;i++) s1+= op (a[i],i==1); S1=S1.SUBSTR (0, V); stringS2=S1.SUBSTR (0,1); S2+="."; S2+=S1.SUBSTR (1); printf ("%s\n", S2.c_str ()); return 0;}//This code can run properly on vs2012

Number of digit required	Time Use (MS)
100	0
1000	31
5000	109
9500	265
10000	297
20000	873
50000	4446
100000	16365
1000000	About 20min

"C + +" Calculate The exact number of E.