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按下面的近似公式计算：
e=1+1/1!+1/2!+1/3!+1/4!+...+1/n!+...使误差小于给定的a.
手头上有一个：
/* Ch24LnBE.c
*
* LongBigE - compute e to ten thousand places, using long int
*
* Copyright (C) 1999 Ian D. K. Kelly,
* idkk Consultancy Ltd.
* Macmillan Computer Publishing
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; either version 2 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 675 Mass Ave, Cambridge, MA 02139, USA.
*
* Ian Kelly may be contacted at idkk@idkk.com
*
*/
/* The comments have been removed from this code. You will find it */
/* instructive to construct comments about it - both the techniques */
/* and the specific values. For a sample set of comments see the file */
/* Ch24BigE.c */
#include <stdio.h>
#include <limits.h>
#define ILIMIT 3300
#if (LONG_MAX>(ILIMIT*100000))
#define NUMBER_BASE 100000
#define DIGS_IN_BASE 5
#else
#define NUMBER_BASE 10
#define DIGS_IN_BASE 1
#endif
#if (DIGS_IN_BASE==5)
#define NUMBER_WIDTH 2001
#else
#define NUMBER_WIDTH 10001
#endif
int LongDivide ( long int * Numerator, long int * Denominator, long int * Answer );
int LongAdd ( long int * iFirst, long int * iSecond, long int * iAnswer );
int main(int argc, char * argv[])
{
long int aResult[NUMBER_WIDTH+1];
long int aWork[NUMBER_WIDTH+1];
long int i;
int j;
aResult[0] = NUMBER_WIDTH;
aResult[1] = 1;
aWork[0] = NUMBER_WIDTH;
aWork[1] = 1;
for (i=2;(i<=NUMBER_WIDTH);i++)
{
aResult[i] = 0;
aWork[i] = 0;
}
for (i=1;(i<ILIMIT);i++)
{
j = LongDivide ( aWork, &i, aWork );
if (j<=0)
{
i = ILIMIT + 1;
}
LongAdd ( aWork, aResult, aResult );
}
printf("e = ");
printf("%1.1d.",aResult[1]);
for (i=2;(i<NUMBER_WIDTH);i++)
{
#if (DIGS_IN_BASE==5)
printf("%5.5d ",aResult[i]);
#endif
#if (DIGS_IN_BASE==1)
printf("%1.1d",aResult[i]);
if ((i % 5)==1)
printf(" ");
#endif
if ((i % 10)==0)
{
printf("\n");
if ((i % 100)==0)
{
printf("\n");
}
}
}
printf("\n");
return 1;
}
int LongDivide ( long int * Numerator, long int * Denominator, long int * Answer )
{
int iStatus = 0;
int j = 0;
long int k = 0;
long int m = 0;
long int d = 0;
d = *Denominator;
Answer[0] = Numerator[0];
for (j=1;(j<=Numerator[0]);j++)
{
if (k>0)
m = Numerator[j] + (k * NUMBER_BASE);
else
m = Numerator[j];
Answer[j] = m / d;
if (Answer[j] !=0)
iStatus = 1;
k = m % d;
}
return iStatus;
}
int LongAdd ( long int * iFirst, long int * iSecond, long int * iAnswer )
{
int iStatus = 0;
int i = 0;
int j = 0;
long int k = 0;
long int m = 0;
i = iFirst[0];
k = 0;
for (j=i;(j>0);j--)
{
if (iFirst[j]!=0)
iStatus = 1;
m = iFirst[j] + iSecond[j] + k;
k = 0;
while (m>=NUMBER_BASE)
{
m -= NUMBER_BASE;
k += 1;
}
iAnswer[j] = m;
}
return iStatus;
}
# include "stdio.h"
main()
{
int i=1,factorial=1,n=0,j;
double error,e=1,c;
printf("请输入误差的值：\n");
scanf("%lf",&error);
if(error>1)
{
printf("误差值太大，请输入不大于1的值:\n");
scanf("%lf",&error);
}
do
{
factorial=factorial*i;
i++;
n++;
}
while(factorial<=(1/error));
printf("n的值是：%d\n",n);
for(i=1;i<=n;i++)
{
factorial=1;
for(j=1;j<=i;j++)
{
factorial=factorial*j;
if(j==i)
c=1/factorial;
}
e=e+c;
}
printf("c=%lf\n",c);
printf("%d",factorial);
}
