COJ 1019 高精度阶乘 S=1！+2！+3！+…+n！（n≤50）可以看作是 1067的深化，但是我不明白为什么我要用long long 来定义数组，好像有点浪费

COJ 1019 高精度阶乘 S=1！+2！+3！+…+n！（n≤50）可以看作是 1067的深化，但是我不明白为什么我要用long long 来定义数组，好像有点浪费

因为要反复求解某数的阶乘，所以求解之前要记得使用memset将数组清零

#include  < cstdlib >
#include  < iostream >

using   namespace  std;

int  main()
{
     long   long  a[ 30000 ],b[ 30000 ];
     void  jiecheng( int  n,  long   long   * a);
     int  n,i,j,carry; 
    cin >> n;
    memset(a, 0 , sizeof (a));
    jiecheng(n,a);
     for (i = n - 1 ;i > 0 ;i -- )
    {
        carry = 0 ;memset(b, 0 , sizeof (b));
        jiecheng(i,b);
         for (j = 1 ;j <= a[ 0 ];j ++ )
        {
            a[j] += b[j] + carry;
            carry  = a[j] / 10 ;
            a[j] = a[j] % 10 ;
        }
         while (carry)
        {
            a[ 0 ] ++ ;
            a[j] = carry % 10 ;
            carry /= 10 ;
            j ++ ;
        }
    }
     for (i = a[ 0 ];i > 0 ;i -- )
        cout << a[i];
    cout << endl; 
    
    system( " PAUSE " );
     return   0 ;
}

void  jiecheng( int  n,  long   long   * a)
{
     long   long  carry,i;
     int  j;    
   
     for (i = 1 ;i <= n;i ++ )
    {
        carry = 0 ;
         if (i == 1 )
        {    
            a[ 0 ] = 1 ;a[ 1 ] = 1 ;
        }
         else
        {
            for (j = 1 ;j <= a[ 0 ];j ++ ) 
           {
                a[j] = i * a[j] + carry;
                carry  =  a[j] / 10 ;
                a[j] = a[j] % 10 ;                
           }
            while (carry)
            {
                a[ 0 ] ++ ;
                a[j] = carry % 10 ;
                carry /= 10 ;
                j ++ ;
            }
        }
    }
     /* for(i=a[0];i>0;i--)
        cout<<a[i];
    cout<<endl; */
}