The number of digits and exact values of a large number of factorial calculations "go"

Source: http://www.cnblogs.com/stonehat/p/3603267.html

Here, worship the original author hehe

We know that the number of bits of integer n is calculated as: log10 (n) +1

So the number of n! is log10 (n!) +1

If the specific value of n! is required, the calculation will be very slow for a large n (for example, n=1000000), and if only the number of digits of the factorial is obtained, the Stirling (Stirling) formula can be used to solve

Sterling (Stirling) formula:

So the number of n! is to seek log10 ((2*pi*n) ^1/2* (n/e) ^n) +1

namely  1/2*log10 (2*pi*n) +n*log10 (n/e) +1

So using the following code to calculate the number of factorial digits, it will be very fast

1 #definePI 3.1415926542 #defineE 2.718281828463 intLintN)4 {5     ints=1;6     if(n>3)7S=LOG10 (2*PI*N)/2+N*LOG10 (n/e) +1;8     returns;9}

If you want to calculate the exact value of a factorial, you can use the following code.

1 /*2 function Function: calculates and outputs the factorial of n3 return Value: The number of digits of the factorial result4 Note:5 This program directly output n! results, need to return the results please keep long a[]6 need math.h7 */8 9 intFactorial (intN)Ten { One     Longa[10000]; A     inti,j,l,c,m=0, W; -a[0]=1;  -      for(i=1; i<=n;i++) the     {  -C=0;  -          for(j=0; j<=m;j++) -         {  +a[j]=a[j]*i+C; -c=a[j]/10000;  +a[j]=a[j]%10000;  A         }  at         if(c>0) {m++;a[m]=C;}  -     }  -  -w=m*4+LOG10 (A[m]) +1; -printf"\n%ld", A[m]);  -      for(i=m-1; i>=0; i--) printf ("%4.4ld", A[i]); in     returnW; -}

The number of digits and exact values of a large number of factorial calculations "go"