N! HDU 1042

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60244    Accepted Submission(s): 17166

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

 

 #include <iostream>
 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<stdlib.h>
 using namespace std;
 int a[];
 int main()
 {
     int n;
     int i,j;
     while(~scanf("%d",&n))
     {
         if(n==)
         {
             printf("1\n");
             continue;
         }
         memset(a,,sizeof(a));
         a[]=;
         for(i=;i<=n;i++)
         {
             for(j=;j<;j++)
             a[j]*=i;
             for(j=;j<;j++)
             {
                 a[j+]+=a[j]/;
                 a[j]%=;
             }
         }
         for(i=;i>=;i--)
         {
             if(a[i]!=)
             break;
         }
         printf("%d",a[i--]);
         for(;i>=;i--)
         printf("%05d",a[i]);
         printf("\n");
     }
     return ;
 }
 -----------------------------------------------
 #include <iostream>
 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<stdlib.h>
 using namespace std;
 int a[];
 int main()
 {
     //freopen("in.txt","r",stdin);
     int n;
     int i,j,temp;
     while(~scanf("%d",&n))
     {
         memset(a,,sizeof(a));
         a[]=;
         int count=;
         for(i=;i<=n;i++)
         {
             int k=;
             for(j=;j<count;j++)
             {
                 temp = a[j]*i+k;
                 a[j]=temp%;
                 k=temp/;
             }
             while(k)
             {
                 a[count++]=k%;
                 k/=;
             }
         }
         for(i=;i>=;i--)
         {
             if(a[i]!=)
             break;
         }
         for(;i>=;i--)
         printf("%d",a[i]);
         printf("\n");
     }
     return ;
 }

模板：

 #include <stdio.h>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>
 #include <assert.h>
 #include <ctype.h>
 #include <map>
 #include <string>
 #include <set>
 #include <bitset>
 #include <utility>
 #include <algorithm>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <iostream>
 #include <fstream>
 #include <list>
 using  namespace  std;

 const  int MAXL = ;
 struct  BigNum
 {
     int  num[MAXL];
     int  len;
     BigNum()
     {
         memset(num,,sizeof(num));
     }
 };

 //高精度比较 a > b return 1, a == b return 0; a < b return -1;
 int  Comp(BigNum &a, BigNum &b)
 {
     int  i;
     if(a.len != b.len) return (a.len > b.len) ?  : -;
     for(i = a.len-; i >= ; i--)
         if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ?  : -;
     return  ;
 }

 //高精度加法
 BigNum  Add(BigNum &a, BigNum &b)
 {
     BigNum c;
     int  i, len;
     len = (a.len > b.len) ? a.len : b.len;
     memset(c.num, , sizeof(c.num));
     for(i = ; i < len; i++)
     {
         c.num[i] += (a.num[i]+b.num[i]);
         if(c.num[i] >= )
         {
             c.num[i+]++;
             c.num[i] -= ;
         }
     }
     if(c.num[len])
         len++;
     c.len = len;
     return  c;
 }
 //高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析
 //如果b很大可以考虑把b看成高精度
 BigNum Mul1(BigNum &a, int  &b)
 {
     BigNum c;
     int  i, len;
     len = a.len;
     memset(c.num, , sizeof(c.num));
     //乘以0，直接返回0
     if(b == )
     {
         c.len = ;
         return  c;
     }
     for(i = ; i < len; i++)
     {
         c.num[i] += (a.num[i]*b);
         if(c.num[i] >= )
         {
             c.num[i+] = c.num[i]/;
             c.num[i] %= ;
         }
     }
     while(c.num[len] > )
     {
         c.num[len+] = c.num[len]/;
         c.num[len++] %= ;
     }
     c.len = len;
     return  c;
 }

 //高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，
 //如果确定不会发生溢出, 可以将里面的while改成if
 BigNum  Mul2(BigNum &a, BigNum &b)
 {
     int i, j, len = ;
     BigNum  c;
     memset(c.num, , sizeof(c.num));
     for(i = ; i < a.len; i++)
     {
         for(j = ; j < b.len; j++)
         {
             c.num[i+j] += (a.num[i]*b.num[j]);
             if(c.num[i+j] >= )
             {
                 c.num[i+j+] += c.num[i+j]/;
                 c.num[i+j] %= ;
             }
         }
     }
     len = a.len+b.len-;
     while(c.num[len-] ==  && len > )
         len--;
     if(c.num[len])
         len++;
     c.len = len;
     return  c;
 }
 void  print(BigNum &a)   //输出大数
 {
     int  i;
     for(i = a.len-; i >= ; i--)
         printf("%d", a.num[i]);
     puts("");
 }

 void Init(BigNum &a, char *s, int &tag)   //将字符串转化为大数
 {
     int  i = , j = strlen(s);
     if(s[] == '-')
     {
         j--;
         i++;
         tag *= -;
     }
     a.len = j;
     for(; s[i] != '\0'; i++, j--)
         a.num[j-] = s[i]-'';
 }

 int main(void)
 {
     int n;
     while(~scanf("%d",&n))
     {
         BigNum a;
         int tag=;
         Init(a,"",tag);
         for(int i=;i<=n;i++)
         {
             a=Mul1(a,i);
         }
         print(a);
     }
     return ;
 }