我需要在C#中计算n!/(n-r)!r!
。对于小数字,使用阶乘函数很容易计算,但是当数字变得像100那样大时,它就不起作用了。有没有其他方法可以计算更大数字的组合?
答案 0 :(得分:16)
首先,我注意到你正在尝试计算二项式系数,所以我们称之为。
以下是一些进行计算的方法。如果你使用BigInteger,你不必担心溢出:
方法一:使用factorial:
static BigInteger Factorial(BigInteger n)
{
BigInteger f = 1;
for (BigInteger i = 2; i <= n; ++i)
f = f * i;
return f;
}
static BigInteger BinomialCoefficient(BigInteger n, BigInteger k)
{
return Factorial(n) / (Factorial(n-k) * Factorial(k));
}
方法二:使用递归:
static BigInteger BinomialCoefficient(BigInteger n, BigInteger k)
{
if (n == 0) return 1;
if (k == 0) return 0;
return BinomialCoefficient(n-1, k-1) + BinomialCoefficient(n-1, k)
}
除非你 memoize 结果,否则这个方法并不快。
方法三:更加巧妙地减少乘法次数,并尽早划分。这使数字变小:
static BigInteger BinomialCoefficient(BigInteger n, BigInteger k)
{
// (n C k) and (n C (n-k)) are the same, so pick the smaller as k:
if (k > n - k) k = n - k;
BigInteger result = 1;
for (BigInteger i = 1; i <= k; ++i)
{
result *= n - k + i;
result /= i;
}
return result;
}
因此,例如,如果你是计算(6 C 3),而不是计算(6 x 5 x 4 x 3 x 2 x 1)/((3 x 2 x 1)x(3 x 2 x 1)) ,你计算(((4/1)* 5)/ 2)* 6)/ 3,如果可能的话,它会使数字变小。
答案 1 :(得分:4)
按照埃里克的说法,尽早划分有很大帮助,你可以通过从高到低来改善它。这样,只要endresult适合Long,您就可以计算任何结果。这是我使用的代码(为Java道歉,但转换应该很容易):
public static long binomialCoefficient(int n, int k) {
// take the lowest possible k to reduce computing using: n over k = n over (n-k)
k = java.lang.Math.min( k, n - k );
// holds the high number: fi. (1000 over 990) holds 991..1000
long highnumber[] = new long[k];
for (int i = 0; i < k; i++)
highnumber[i] = n - i; // the high number first order is important
// holds the dividers: fi. (1000 over 990) holds 2..10
int dividers[] = new int[k - 1];
for (int i = 0; i < k - 1; i++)
dividers[i] = k - i;
// for every divider there always exists a highnumber that can be divided by
// this, the number of highnumbers being a sequence that equals the number of
// dividers. Thus, the only trick needed is to divide in reverse order, so
// divide the highest divider first trying it on the highest highnumber first.
// That way you do not need to do any tricks with primes.
for (int divider: dividers)
for (int i = 0; i < k; i++)
if (highnumber[i] % divider == 0) {
highnumber[i] /= divider;
break;
}
// multiply remainder of highnumbers
long result = 1;
for (long high : highnumber)
result *= high;
return result;
}
答案 2 :(得分:0)
对于.net 4.0及更大版本,请使用类BigInteger而不是int / long
对于其他.net使用自定义大号类,例如来自IntX:http://www.codeplex.com/IntX/
答案 3 :(得分:0)
我认为这将是有效的 它是O(k)
注意n! / r! r!只是取消n的最后一个r 所以7 37 x 6 x 5 x 4 x 3 x 2 x 1
over
4 x 3 x 2 x 1
public static uint BinomialCoeffient(uint n, uint k)
{
if (k > n)
return 0;
uint c = n;
for (uint i = 1; i < k; i++)
{
c *= n - i;
c /= i + 1;
}
return c;
}