【ProjectEuler】ProjectEuler_053（对于1≤n≤100，C(n,r)有多少超过100万？）

// Combinatoric selections
//     Problem 53
//     There are exactly ten ways of selecting three from five, 12345:
//
// 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
//
//     In combinatorics, we use the notation, 5C3 = 10.
//
//     In general,
//
//     nCr = n! / r!(n-r)!
//     ,where r  n, n! = n(n1)...321, and 0! = 1.
//     It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
//
//     How many, not necessarily distinct, values of  nCr, for 1  n  100, are greater than one-million?
//
// 题目53:对于1≤n≤100，C(n,r)有多少超过100万？
// 从五个数12345中选出三个数一共有十种方法:
//
//      123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
//
//          在组合数学中我们用5C3 = 10来表示.
//
//          概括来说：
//
//          nCr = n! / r!(n-r)!
//          ,其中r  n, n! = n(n1)...321, 并且0! = 1.
//          n = 23时产生第一个超过一百万的数: 23C10 = 1144066.
//
//          对于nCr,  1 <= n <= 100，有多少超过100万的值？包括重复的在内。
//

#include 
#include 
#include 
#include 

#include 

using namespace std;

// 打印时间等相关信息
class DetailPrinter
{
public:
    void Start();
    void End();
    DetailPrinter();

private:
    LARGE_INTEGER timeStart;
    LARGE_INTEGER timeEnd;
    LARGE_INTEGER freq;
};

DetailPrinter::DetailPrinter()
{
    QueryPerformanceFrequency(&freq);
}

//************************************
// Method:    Start
// Access:    public
// Describe:  执行每个方法前调用
// Returns:   void
//************************************
void DetailPrinter::Start()
{
    QueryPerformanceCounter(&timeStart);
}

//************************************
// Method:    End
// Access:    public
// Describe:  执行每个方法后调用
// Returns:   void
//************************************
void DetailPrinter::End()
{
    QueryPerformanceCounter(&timeEnd);
    cout << "Total Milliseconds is " << (double)(timeEnd.QuadPart - timeStart.QuadPart) * 1000 / freq.QuadPart << endl;

    const char BEEP_CHAR = '\007';

    cout << endl << "By GodMoon" << endl << __TIMESTAMP__ << BEEP_CHAR << endl;

    system("pause");
}

/*************************解题开始*********************************/

/*
 * 此题因为数字很大，而且只需要比较是否大于1000000，因此采用浮点数做组合的计算结果。
 */
 
//***********************************
// Method:    CombinationDouble
// Access:    public 
// Describe:  计算组合，返回值为double类型
// Parameter: UINT32 n
// Parameter: UINT32 r
// Returns:   double
//************************************
double CombinationDouble(UINT32 n, UINT32 r)
{
    // C(n,r)=n!/r!/(n-r)!=(r+1)*(r+2)*...*n/1/2/.../(n-r)
    double result=1;

    for (UINT32 i=r+1;i<=n;++i)
    {
        result*=i;
    }

    UINT32 d=n-r;
    for (UINT32 i=2;i<=d;++i)
    {
        result/=i;
    }

    return result;
}

void TestFun1()
{

    cout << "TestFun1 OK!" << endl;
}

void F1()
{
    cout << "void F1()" << endl;

    // TestFun1();

    DetailPrinter detailPrinter;
    detailPrinter.Start();

    /*********************************算法开始*******************************/
    const UINT32 BEGIN_NUM = 1;
    const UINT32 END_NUM = 100;

    const UINT32 LIMIT_NUM = 1000000;

    UINT32 count = 0;
    UINT32 r;

    for(UINT32 n = BEGIN_NUM; n <= END_NUM; ++n)
    {
        for(r = BEGIN_NUM; r <= n; ++r)
        {
            if(CombinationDouble(n, r) >= LIMIT_NUM)
            {
                ++count;
            }
        }
    }

    cout << "n包含于[" << BEGIN_NUM << "," << END_NUM << "]，C(n,r)超过" << LIMIT_NUM << "的数一共有" << count << "个" << endl;

    /*********************************算法结束*******************************/

    detailPrinter.End();
}

//主函数
int main()
{
    F1();
    return 0;
}

/*
void F1()
n包含于[1,100]，C(n,r)超过1000000的数一共有4075个
Total Milliseconds is 13.93

By GodMoon
Fri Apr  5 11:28:00 2013
*/