你的公式完全错误,正确的应该是fact(n)/fact(r)/fact(n-r)
,但这种计算方式效率非常低下。
请参见快速计算多分类组合数,尤其是我的评论。(哦,还有,请重新打开那个问题,这样我才能正确回答)
单一拆分情况实际上非常容易处理:
unsigned nChoosek( unsigned n, unsigned k )
{
if (k > n) return 0;
if (k * 2 > n) k = n-k;
if (k == 0) return 1;
int result = n;
for( int i = 2; i <= k; ++i ) {
result *= (n-i+1);
result /= i;
}
return result;
}
演示:http://ideone.com/aDJXNO
如果结果不适合,您可以计算对数的总和,并以double形式获得近似组合数。或者使用任意精度整数库。
我将我的解决方案放在这篇与之密切相关的问题中,因为ideone.com最近丢失了代码片段,而那个问题仍然关闭以防止新答案的提交。
#include <utility>
#include <vector>
std::vector< std::pair<int, int> > factor_table;
void fill_sieve( int n )
{
factor_table.resize(n+1);
for( int i = 1; i <= n; ++i )
factor_table[i] = std::pair<int, int>(i, 1);
for( int j = 2, j2 = 4; j2 <= n; (j2 += j), (j2 += ++j) ) {
if (factor_table[j].second == 1) {
int i = j;
int ij = j2;
while (ij <= n) {
factor_table[ij] = std::pair<int, int>(j, i);
++i;
ij += j;
}
}
}
}
std::vector<unsigned> powers;
template<int dir>
void factor( int num )
{
while (num != 1) {
powers[factor_table[num].first] += dir;
num = factor_table[num].second;
}
}
template<unsigned N>
void calc_combinations(unsigned (&bin_sizes)[N])
{
using std::swap;
powers.resize(0);
if (N < 2) return;
unsigned& largest = bin_sizes[0];
size_t sum = largest;
for( int bin = 1; bin < N; ++bin ) {
unsigned& this_bin = bin_sizes[bin];
sum += this_bin;
if (this_bin > largest) swap(this_bin, largest);
}
fill_sieve(sum);
powers.resize(sum+1);
for( unsigned i = largest + 1; i <= sum; ++i ) factor<+1>(i);
for( unsigned bin = 1; bin < N; ++bin )
for( unsigned j = 2; j <= bin_sizes[bin]; ++j ) factor<-1>(j);
}
#include <iostream>
#include <cmath>
int main(void)
{
unsigned bin_sizes[] = { 8, 1, 18, 19, 10, 10, 7, 18, 7, 2, 16, 8, 5, 8, 2, 3, 19, 19, 12, 1, 5, 7, 16, 0, 1, 3, 13, 15, 13, 9, 11, 6, 15, 4, 14, 4, 7, 13, 16, 2, 19, 16, 10, 9, 9, 6, 10, 10, 16, 16 };
calc_combinations(bin_sizes);
char* sep = "";
for( unsigned i = 0; i < powers.size(); ++i ) {
if (powers[i]) {
std::cout << sep << i;
sep = " * ";
if (powers[i] > 1)
std::cout << "**" << powers[i];
}
}
std::cout << "\n\n";
}
fact(n-1)
(它们相互抵消了)。 - Ben Voigt