N!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9907 Accepted Submission(s): 2330
Problem 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
Author
JGShining(极光炫影)
#include<stdio.h>
void main(void)
{
int b,c,e,s,x,d,m,n,i;
int a[10000];
while(scanf("%d",&n)!=EOF)
{
m=1; a[0]=1;
for(x=2;x<=n;x++)
{
for(i=0;i<m;i++)
a[i]=a[i]*x;
for(d=0;d<m;d++)
{
if(a[d]>100000)
{
if(d<m-1)
a[d+1]=a[d]/100000+a[d+1];
else{
m++;
a[d+1]=a[d]/100000;
}
a[d]=a[d]%100000;
}
}
void main(void)
{
int b,c,e,s,x,d,m,n,i;
int a[10000];
while(scanf("%d",&n)!=EOF)
{
m=1; a[0]=1;
for(x=2;x<=n;x++)
{
for(i=0;i<m;i++)
a[i]=a[i]*x;
for(d=0;d<m;d++)
{
if(a[d]>100000)
{
if(d<m-1)
a[d+1]=a[d]/100000+a[d+1];
else{
m++;
a[d+1]=a[d]/100000;
}
a[d]=a[d]%100000;
}
}
}
printf("%d",a[m-1]);
for(i=m-2;i>=0;i--)
printf("%05d",a[i]);
printf("/n");
}
}
printf("%d",a[m-1]);
for(i=m-2;i>=0;i--)
printf("%05d",a[i]);
printf("/n");
}
}