快速组合数
递推公式很简单：
C(n,k+1) = C(n,k) * (n-k) / (k + 1)
方法很暴力
经测，C（2000，1000）可以求出，C（2000，0）到C（2000，2000）所用时间仅需0.2s
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <limits>
#include <cstdlib>
using namespace std;
const int MAXD = 100, DIG = 9, BASE = 1000000000;
const unsigned long long BOUND = numeric_limits <unsigned long long> :: max () - (unsigned long long) BASE * BASE;
class bignum
{
private:
int digits[MAXD];
int D;
public:
friend ostream &operator<<(ostream &out,bignum &c);
inline void trim()
{
while(D > 1 && digits[D-1] == 0 )
D--;
}
inline void dealint(long long x)
{
memset(digits,0,sizeof(digits));
D = 0;
do
{
digits[D++] = x % BASE;
x /= BASE;
}
while(x > 0);
}
inline void dealstr(char *s)
{
memset(digits,0,sizeof(digits));
int len = strlen(s),first = (len + DIG -1)%DIG + 1;
D = (len+DIG-1)/DIG;
for(int i = 0; i < first; i++)
digits[D-1] = digits[D-1]*10 + s[i] - '0';
for(int i = first, d = D-2; i < len; i+=DIG,d--)
for(int j = i; j < i+DIG; j++)
digits[d] = digits[d]*10 + s[j]-'0';
trim();
}
inline char *print()
{
trim();
char *cdigits = new char[DIG * D + 1];
int pos = 0,d = digits[D-1];
do
{
cdigits[pos++] = d % 10 + '0';
d/=10;
}
while(d > 0);
reverse(cdigits,cdigits+pos);
for(int i = D - 2; i >= 0; i--,pos += DIG)
for(int j = DIG-1,t = digits[i]; j >= 0; j--)
{
cdigits[pos+j] = t%10 + '0';
t /= 10;
}
cdigits[pos] = '\0';
return cdigits;
}
bignum()
{
dealint(0);
}
bignum(long long x)
{
dealint(x);
}
bignum(int x)
{
dealint(x);
}
bignum(char *s)
{
dealstr(s);
}
inline bool operator < (const bignum &o) const
{
if(D != o.D)
return D < o.D;
for(int i = D-1; i>=0; i--)
if(digits[i] != o.digits[i])
return digits[i] < o.digits[i];
return false; //equal
}
bool operator > (const bignum & o)const
{
return o < *this;
}
bool operator <= (const bignum & o)const
{
return !(o < *this);
}
bool operator >= (const bignum & o)const
{
return !(*this < o);
}
bool operator != (const bignum & o)const
{
return o < *this || *this < o;
}
bool operator == (const bignum & o)const
{
return !(o < *this) && !(*this < o);
}
bignum &operator++()
{
*this = *this + 1;
return *this;
}
bignum operator ++(int)
{
bignum old = *this;
++(*this);
return old;
}
inline bignum operator << (int p) const
{
bignum temp;
temp.D = D + p;
for (int i = 0; i < D; i++)
temp.digits [i + p] = digits [i];
for (int i = 0; i < p; i++)
temp.digits [i] = 0;
return temp;
}
inline bignum operator >> (int p) const
{
bignum temp;
temp.D = D - p;
for (int i = 0; i < D - p; i++)
temp.digits [i] = digits [i + p];
for (int i = D - p; i < D; i++)
temp.digits [i] = 0;
return temp;
}
bignum &operator += (const bignum &b)
{
*this = *this + b;
return *this;
}
bignum &operator -= (const bignum &b)
{
*this = *this - b;
return *this;
}
bignum &operator *= (const bignum &b)
{
*this = *this * b;
return *this;
}
bignum &operator /= (const bignum &b)
{
*this = *this / b;
return *this;
}
bignum &operator %= (const bignum &b)
{
*this = *this % b;
return *this;
}
inline bignum operator + (const bignum &o) const
{
bignum sum = o;
int carry = 0;
for (sum.D = 0; sum.D < D || carry > 0; sum.D++)
{
sum.digits [sum.D] += (sum.D < D ? digits [sum.D] : 0) + carry;
if (sum.digits [sum.D] >= BASE)
{
sum.digits [sum.D] -= BASE;
carry = 1;
}
else
carry = 0;
}
sum.D = max (sum.D, o.D);
sum.trim ();
return sum;
}
inline bignum operator - (const bignum &o) const
{
bignum diff = *this;
for (int i = 0, carry = 0; i < o.D || carry > 0; i++)
{
diff.digits [i] -= (i < o.D ? o.digits [i] : 0) + carry;
if (diff.digits [i] < 0)
{
diff.digits [i] += BASE;
carry = 1;
}
else
carry = 0;
}
diff.trim ();
return diff;
}
inline bignum operator * (const bignum &o) const
{
bignum prod = 0;
unsigned long long sum = 0, carry = 0;
for (prod.D = 0; prod.D < D + o.D - 1 || carry > 0; prod.D++)
{
sum = carry % BASE;
carry /= BASE;
for (int j = max (prod.D - o.D + 1, 0); j <= min (D - 1, prod.D); j++)
{
sum += (unsigned long long) digits [j] * o.digits [prod.D - j];
if (sum >= BOUND)
{
carry += sum / BASE;
sum %= BASE;
}
}
carry += sum / BASE;
prod.digits [prod.D] = sum % BASE;
}
prod.trim ();
return prod;
}
inline bignum range (int a, int b) const
{
bignum temp = 0;
temp.D = b - a;
for (int i = 0; i < temp.D; i++)
temp.digits [i] = digits [i + a];
return temp;
}
inline double double_div (const bignum &o) const
{
double val = 0, oval = 0;
int num = 0, onum = 0;
for (int i = D - 1; i >= max (D - 3, 0); i--, num++)
val = val * BASE + digits [i];
for (int i = o.D - 1; i >= max (o.D - 3, 0); i--, onum++)
oval = oval * BASE + o.digits [i];
return val / oval * (D - num > o.D - onum ? BASE : 1);
}
inline pair <bignum, bignum> divmod (const bignum &o) const
{
bignum quot = 0, rem = *this, temp;
for (int i = D - o.D; i >= 0; i--)
{
temp = rem.range (i, rem.D);
int div = (int) temp.double_div (o);
bignum mult = o * div;
while (div > 0 && temp < mult)
{
mult = mult - o;
div--;
}补充：软件开发 , C++ ,
