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Fri, 29 Jun 2007 09:12:00 GMT

哈胖�?/a> 2007-06-29 17:12 发表评论

Thu, 28 Jun 2007 07:58:00 GMT

哈胖�?/a> 2007-06-28 15:58 发表评论

Thu, 28 Jun 2007 07:37:00 GMT

通配�W�匹配属于正则表辑ּ�匚w��的一�U�特�D�情况，它包含两�U�通配字符�Q?�?�?/p>

其中通配�W?* 表示可以匚w��0个或多个��L��的字�W�；

其中通配�W?? 表示可以匚w��0个或1个�Q意字�W��?/p>

如果不想寚w��配的模式进行预处理�Q�可以��用如下递归的wildcard匚w��Q�只不过�q�种匚w��方式比较低效。其实，位�ƈ行算�?bit-parallel)��法是处理通配�W�匹配的一�U�有效技术，我们��另文介�l��?/p>

/*
recursive wildcard matching
* : match zero or more characters;
? : match zero or one character.

return value:
1 : match success.
0 : no match.
*/
int wildcard(const char * pat, const char * str)
{
while(*str && *pat)
{
   if(*pat=='?')
   {
    if(wildcard(pat+1, str)) return 1;
    str++;
    pat++;
   }
   else if(*pat=='*')
   {
    while(*pat=='*' || *pat=='?') pat++;
    if(!*pat) return 1;

    while(*str)
    {
     if(wildcard(pat, str)) return 1;
     str++;
    }
    return 0;
   }
   else
   {
    if(*pat!=*str) return 0;
    str++;
    pat++;
   }
}

if(*str!=0) return 0;

while(*pat=='*' || *pat=='?') pat++;
return !*pat;
}

哈胖�?/a> 2007-06-28 15:37 发表评论

Tue, 02 Jan 2007 11:34:00 GMT

在编�E�中�Q�经帔R��要计��最大公�U�数和最��公倍数。本文介�l�计��他们的�Ҏ��?br>在数��Z��Q�两个数a和b的最大公�U�数��Cؓgcd(a, b)�Q�最��公倍数��Cؓlcm(a, b)�?/p>

计算最大公�U�数的算法是Euclid��法�Q�即所谓的“辗�{盔R��?#8221;�Q?/p>

int gcd( int a, int b)
{
return (b == 0 ) ? a : gcd(b, a % b);
}

该算法的旉��复杂度�ؓO(log(max(a, b)))�Q�因而是相当快的�?/p>

�Ҏ��初等数论的知识：对于��L��的a和b�Q�gcd(a, b) * lcm(a, b)=a*b。据此很�Ҏ��得到计算最��公倍数的方法：

int lcm(int a, int b)
{
return (a*b)/gcd(a, b);
}

另外一个计��lcm的方法来自newsmth的programming版，它只用加减法卛_��完成计算�Q?

int LCM(int a, int b)
{
int x=a, y=b;
int u=b, v=a;

while(x!=y)
{
     if(x>y)
     {
      x-=y;
      v+=u;
     }
     else
     {
      y-=x;
      u+=v;
     }
}

return (u+v)/2;
}

哈胖�?/a> 2007-01-02 19:34 发表评论

TopCoder-srm-283-II-3

Wed, 27 Dec 2006 12:57:00 GMT

Problem Statement
The greatest common divisor (GCD) of two positive integers a and b is the largest integer that evenly divides both a and b. In this problem, you will find the GCD of a positive integer and the factorial of a non-negative integer.

Create a class FactorialGCD with method factGCD which takes an int a and an int b as parameters and returns the GCD of a! (the factorial of a) and b.

#include < utility >
#include < iostream >
using namespace std;

class FactorialGCD
{
public :
int factGCD( int a, int b)
{
  unsigned int gcd = 1 ;

   while (b > 1 )
   {
   unsigned int p;

    for (p = 2 ; p * p <= (unsigned int )b && b % p != 0 ; p ++ ) ;
             if (b % p != 0 ) p = b;

    int n = 0 ;
    for ( ; b % p == 0 ; b /= p) n ++ ;

    int m = 0 ;
    for (unsigned int j = a; j >= p; ) m += (j /= p);

    for ( int i = 0 ; i < min(n, m); i ++ ) gcd *= p;
  }

   return gcd;
}
} ;

int main()
{
FactorialGCD test;

cout << test.factGCD( 5 , 20 ) << endl;

cout << test.factGCD( 7 , 5040 ) << endl;

cout << test.factGCD( 0 , 2425 ) << endl;

cout << test.factGCD( 667024 , 1 ) << endl;

cout << test.factGCD( 4 , 40 ) << endl;

cout << test.factGCD( 2097711064 , 2147483646 ) << endl;

return 1 ;
}

哈胖�?/a> 2006-12-27 20:57 发表评论

�q�制转换

Fri, 22 Dec 2006 12:07:00 GMT

#include < iostream >
using namespace std;
#include < string .h >

void BaseConv1( char * s, int a, char * t, int b)
{
     char * q = t;

     while ( * s)
     {
         int r = 0 ;

         for ( char * p = s; * p; p ++ )
         {
            r = a * r +* p - ' 0 ' ;
             * p = r / b + ' 0 ' ;
            r %= b;
        }

         * q ++= r + ' 0 ' ;

         while ( * s == ' 0 ' ) s ++ ;
    }

     * q = ' \0 ' ;

     for ( char * u = t, * v = q - 1 ; u < v; u ++ , v -- )
     {
         char c =* u;
         * u =* v;
         * v = c;
    }
}

void BaseConv2( const char * s, int a, char * t, int b)
{
     char * p = t;

     for ( * p = ' \0 ' ; * s; s ++ )
     {
         int x =* s - ' 0 ' ;

         for (p = t; * p; p ++ )
         {
            x = a * ( * p - ' 0 ' ) + x;
             * p = x % b + ' 0 ' ;
            x /= b;
        }

         while (x)
         {
             * p ++= x % b + ' 0 ' ;
            x /= b;
        }

         * p = ' \0 ' ;
    }

     for ( char * u = t, * v = p - 1 ; u < v; u ++ , v -- )
     {
         char c =* u;
         * u =* v;
         * v = c;
    }
}

int main()
{
     char s[] = " 987654321014612555422121454541212229852142104542345678901 " ;
     char t[ 4096 ];

    cout << " s= " << s << endl;
    BaseConv2(s, 10 , t, 8 );
    cout << " t= " << t << endl;

    cout << " s= " << s << endl;
    BaseConv1(s, 10 , t, 8 );
    cout << " t= " << t << endl;

     return 1 ;
}

哈胖�?/a> 2006-12-22 20:07 发表评论