hdoj1042(N!，大数乘)

HDOJ1042(N!)(大数乘)

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 67802    Accepted Submission(s): 19420

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
  
  

 

My   solution:

/*2015.10.29*/

#include<stdio.h>
#include<string.h>
int t1[10],t2[40000],t3[40000],n,m;
void mul()
{
	int i,j,k,g;
	k=n;
	for(i=0;i<m;i++)
	{
		k=i;
		for(j=0;j<n;j++)
		{
			t3[k++]+=t1[i]*t2[j];
		}
	}
	for(i=0;i<k;i++)
	{
		if(t3[i]>9)/*t3起转换容器的变量，暂时存放计算结果*/
		{
			g=t3[i]/10;
			t3[i]%=10;
			t3[i+1]+=g;
			if(i==k-1)/*最高位要进位，需要把数组长度再加1*/
			k=k+1; 
		}
		t2[i]=t3[i];
		t3[i]=0;/*一定要把t3数组清空,否则下一次运算时,会再次加上，导致结果出错*/
	}
     n=k;/*更新t2数组长度*/	
}
void  tiaozhen()
{
	int i,j;
	t1[0]+=1;/*每乘一次，待乘数要加1,t1[0]一定要初始化为1*/ 
	for(i=0;i<m;i++)
	{
		if(t1[i]>9)/*逢10进1*/
		{
			j=t1[i]/10;
			t1[i]%=10;
			t1[i+1]+=j;
			if(i==m-1)/*最高位要进位，需要把数组长度再加1*/
			m+=1;
		}
	 } 
}
void Njie(int e)
{
	int i,j;
	for(i=2;i<=e;i++)
	{
		tiaozhen();
		mul();
	 } 
}
void print()
{
	int i;
	for(i=n-1;i>=0;i--)
	printf("%d",t2[i]);
	printf("\n");
}
int main()
{
	int i,e;
	while(scanf("%d",&e)==1)
	{
		if(e>1)
		{
			memset(t1,sizeof(t1));
			memset(t2,sizeof(t2));
			memset(t3,sizeof(t3));
			n=m=1;
			t1[0]=1;
			t2[0]=1;
			Njie(e);
			print();
		}
		else
		printf("1\n");
	}
	return 0;
}