快速计算C(n,r)

在网上见的，引用出处为：http://blog.csdn.net/alexingcool/article/details/7997599

可以在logn内计算出，但是容易溢出。

[cpp] view plain copy
01.#include <iostream>
02.#include <iterator>
03.#include <algorithm>
04.
05.using namespace std;
06.
07.#define MAXSIZE 100
08.unsigned long answer[MAXSIZE];
09.
10.unsigned long long int power(unsigned long base, unsigned long exp)
11.{
12.    unsigned long long int result = 1;
13.    while (exp)
14.    {
15.        if (exp & 0x01)
16.            result *= base;
17.        base *= base;
18.        exp >>= 1;
19.    }
20.
21.    return result;
22.}
23.
24.void cnr(unsigned long n, unsigned long *answer)
25.{
26.    unsigned long long int x = (1 << n) + 1;
27.    unsigned long long int mask = (1 << n) - 1;
28.    unsigned long long int result;
29.
30.    result = power(x, n);
31.    for (unsigned long i = 0; i <= n; i++, result >>= n)
32.    {
33.        cout << result << endl;
34.        answer[i] = result & mask;
35.        cout << "answer[" << i << "] = " << answer[i] << endl;
36.    }
37.}
38.
39.void main()
40.{
41.    int number = 4;
42.    cnr(number, answer);
43.    copy(answer, answer + number, ostream_iterator<int>(cout, " "));
44.} 

当然，可以用动归，慢点，n²：
利用公式C(n,r)=C(n-1,r-1)+C(n-1,r)来化简求解：

01.#include <iostream>
02.
03.using namespace std;
04.
05.#define MAXSIZE 100
06.
07.unsigned long cnr(int n, int r)
08.{
09.    unsigned long c[MAXSIZE];
10.    int i, j;
11.    for (i = 0; i <= r; i++)
12.        c[i] = 1;
13.    for (i = 1; i <= n - r; i++)
14.        for (j = 1; j <= r; j++)
15.            c[j] += c[j - 1];
16.    return c[r];
17.}
18.
19.void main()
20.{
21.    int result = cnr(10, 2);
22.    cout << "result = " << result << endl;
23.}  

还可以用质因数分解，不易溢出。