HDU 1042 N! (large number factorial, or Java Dafa Good, C can not miss!!!)

N!Time limit:10000/5000 MS (java/others) Memory limit:262144/262144 K (java/others)
Total submission (s): 73270 Accepted Submission (s): 21210


Problem Descriptiongiven an integer n (0≤n≤10000), your task is to calculate N!

Inputone N in one line, process to the end of file.

Outputfor each n, output N! In one line.

Sample Input

123

Sample Output

126

Authorjgshining (Aurora Dazzle)
original title link: http://acm.hdu.edu.cn/showproblem.php?pid=1042
big number! or Java Dafa good, Rujia Purple book has a factorial example, but time-out!!! See the code for specific reasons!
AC Code 1 (Java)

Import Java.math.bigdecimal;import Java.util.scanner;public class Main {public static void main (string[] args) {Scanner s c = new Scanner (system.in), while (Sc.hasnext ()) {int n = sc.nextint (); System.out.println (Fun (n));}} public static BigDecimal Fun (int n) {BigDecimal s = new BigDecimal (1), for (int i = 1; I <= n; i++) {BigDecimal a = new BigDecimal (i); s = s.multiply (a);} return s;}}

AC Code 2 (c analogue)

#include <stdio.h> #include <string.h> #define MAXN 50000int f[maxn];int Main () {    int i,j,n;    while (scanf ("%d", &n)!=eof)    {        memset (f,0,sizeof (f));        F[0]=1;        int count=1;        for (I=1; i<=n; i++)        {            int c=0;//carry for            (j=0; j<count; j + +)//factorial bits            {                int s=f[j]*i+c;                f[j]=s%10;                C=S/10;            }            while (c)            {                f[count++]=c%10;                c/=10;            }        }        for (j=maxn-1; j>=0; j--)            if (f[j]) break;        for (i=j; i>=0; i--)            printf ("%d", F[i]);        printf ("\ n");    }    return 0;}

Timeout Code:

#include <stdio.h> #include <string.h> #define MAXN 50000int f[maxn];int Main () {    int i,j,n;    while (scanf ("%d", &n)!=eof)    {        memset (f,0,sizeof (f));        F[0]=1;        for (i=2; i<=n; i++)        {            int c=0;            for (j=0; j<maxn; j + +)//length, each time to maximum, resulting in timeouts!!!!            {                int s=f[j]*i+c;                f[j]=s%10;                C=S/10;            }        }        for (j=maxn-1; j>=0; j--)            if (f[j]) break;        for (i=j; i>=0; i--)            printf ("%d", F[i]);        printf ("\ n");    }    return 0;}

HDU 1042 N! (large number factorial, or Java Dafa Good, C can not miss!!!)