/*Crawling in process... Crawling failed Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!
Input
One N in one line, process to the end of file.
Output
For each N, output N! in one line.
Sample Input
1
2
3
Sample Output
1
2
6
*
#include<stdio.h>
#include<string.h>
const int max= 50000;
int f[max];
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<max;j++) //存到數組裏。
{
int s=f[j]*i+c;
f[j]=s%10;
c=s/10;
}
}
for(j=max-1;j>=0;j--) //輸出那個數組。
if(f[j])
break;
for(i=j;i>=0;i--)
printf("%d",f[i]);
printf("\n");
}
return 0;
}*/
#include <stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
#define N 10001 //計算N的階乘,修改N的定義可計算10000以內任意數的階乘
long a[N],n,i,c,m;
int main()
{
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
a[0]=1;
m=0;
for(;n>1;(a[i]=c)>0?m++:0,n--)//m是最高位
{
for(c=0,i=0;i<=m;i++)//c爲累加保存的進位
{
a[i]=(c+=a[i]*n)%10000;
c/=10000;
}
}
for(printf("%d",a[m]);--m>=0;)
printf("%04d",a[m]);
printf("\n");
}
return 0;
點擊打開鏈接
#include<stdio.h>
#include<string.h>
#define N 50000 //計算N的階乘,最多隻能算到9999
long a[N],n,i,c,m;
int main()
{
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
a[0]=1;
m=0;
for ( ;n>1; (a[i]=c)>0?m++:0,n--)
{
for (c=0,i=0; i<=m;i++)
{
a[i]= (c+=a[i]*n)%10000;
c/=10000;
}
}
printf("%d",a[m]);
for( m--;m>=0;m--)
printf("%04d",a[m]);
printf("\n");
}
return 0;
}