【ProjectEuler】ProjectEuler_050（找到100萬以内最多連續素數的和，它也同時是個素數）

#pragma once

#include <windows.h>
#include <vector>
#include <set>

using namespace std;

class MoonMath
{
public:
    MoonMath(void);
    ~MoonMath(void);

    //************************************
    // Method:    IsInt
    // Access:    public
    // Describe:  判斷double值在epsilon的範圍内是否很接近整數
    //            如1.00005在epsilon為0.00005以上就很接近整數
    // Parameter: double doubleValue    要判斷的double值
    // Parameter: double epsilon        判斷的精度，0 < epsilon < 0.5
    // Parameter: INT32 & intValue      如果接近，傳回最接近的整數值
    // Returns:   bool                  接近傳回true，否則傳回false
    //************************************
    static bool IsInt(double doubleValue, double epsilon, INT32 &intValue);

    //************************************
    // Method:    Sign
    // Access:    public
    // Describe:  擷取value的符号
    // Parameter: T value   要擷取符号的值
    // Returns:   INT32     正數、0和負數分别傳回1、0和-1
    //************************************
    template <typename T>
    static INT32 Sign(T value);

    const static UINT32 MIN_PRIMER = 2;     // 最小的素數

    //************************************
    // Method:    IsPrimer
    // Access:    public
    // Describe:  判斷一個數是否是素數
    // Parameter: UINT32 num    要判斷的數
    // Returns:   bool          是素數傳回true，否則傳回false
    //************************************
    static bool IsPrimer(UINT32 num);

    //************************************
    // Method:    IsIntegerSquare
    // Access:    public static
    // Describe:  判斷給定的數開平方後是否為整數
    // Parameter: UINT32 num
    // Returns:   bool
    //************************************
    static bool IsIntegerSquare(UINT32 num);

    //************************************
    // Method:    GetDiffPrimerFactorNum
    // Access:    public static
    // Describe:  擷取num所有的不同質因數
    // Parameter: UINT32 num
    // Returns:   set<UINT32>
    //************************************
    static set<UINT32> MoonMath::GetDiffPrimerFactorNum(UINT32 num);
};
           

#include "MoonMath.h"
#include <cmath>

MoonMath::MoonMath(void)
{
}


MoonMath::~MoonMath(void)
{
}

template <typename T>
INT32 MoonMath::Sign(T value)
{
    if(value > 0)
    {
        return 1;
    }
    else if(value == 0)
    {
        return 0;
    }
    else
    {
        return -1;
    }
}

bool MoonMath::IsInt(double doubleValue, double epsilon, INT32 &intValue)
{
    if(epsilon > 0.5 || epsilon < 0)
    {
        return false;
    }

    if(INT32(doubleValue + epsilon) == INT32(doubleValue - epsilon))
    {
        return false;
    }

    INT32 value = INT32(doubleValue);

    intValue = (fabs(doubleValue - value) > 0.5) ? (value + MoonMath::Sign(doubleValue)) : (value) ;
    return true;
}

bool MoonMath::IsPrimer(UINT32 num)
{
    if(num < MIN_PRIMER)
    {
        return false;
    }

    UINT32 sqrtOfNum = (UINT32)sqrt((double)num); // num的2次方

    // 從MIN_PRIMER到sqrt(num)，如果任何數都不能被num整除，num是素數，否則不是
    for(UINT32 i = MIN_PRIMER; i <= sqrtOfNum; ++i)
    {
        if(num % i == 0)
        {
            return false;
        }
    }

    return true;
}

bool MoonMath::IsIntegerSquare(UINT32 num)
{
    UINT32 qurtNum = (UINT32)sqrt((double)num);

    return (qurtNum * qurtNum) == num;
}

set<UINT32> MoonMath::GetDiffPrimerFactorNum(UINT32 num)
{
    UINT32 halfNum = num / 2;
    set<UINT32> factors;

    for(UINT32 i = 2; i <= halfNum; ++i)
    {
        if(!MoonMath::IsPrimer(i))
        {
            continue;
        }

        if(num % i == 0)
        {
            factors.insert(i);

            while(num % i == 0)
            {
                num /= i;
            }
        }
    }

    return factors;
}
           

// Consecutive prime sum
//     Problem 50
//     The prime 41, can be written as the sum of six consecutive primes:
//
// 41 = 2 + 3 + 5 + 7 + 11 + 13
//     This is the longest sum of consecutive primes that adds to a prime below one-hundred.
//
//     The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
//
//     Which prime, below one-million, can be written as the sum of the most consecutive primes?

#include <iostream>
#include <windows.h>
#include <ctime>
#include <assert.h>

#include <vector>
#include <MoonMath.h>

using namespace std;

// 列印時間等相關資訊
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:  執行每個方法前調用
// Returns:   void
//************************************
void DetailPrinter::Start()
{
    QueryPerformanceCounter(&timeStart);
}

//************************************
// Method:    End
// Access:    public
// Describe:  執行每個方法後調用
// 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");
}

/*************************解題開始*********************************/

//************************************
// Method:    MakePrimerVector
// Access:    public
// Describe:  擷取[0,maxNum]的所有素數，存入primers
// Parameter: UINT32 maxNum             要找的最大值
// Parameter: vector<UINT32> & primers  輸出素數數組，素數由小到大排列
// Returns:   void
//************************************
void MakePrimerVector(UINT32 maxNum, vector<UINT32> &primers)
{
    for(UINT32 num = MoonMath::MIN_PRIMER; num <= maxNum; ++num)
    {
        if(MoonMath::IsPrimer(num))
        {
            primers.push_back(num);
        }
    }
}


//************************************
// Method:    GetSum
// Access:    public
// Describe:  計算疊代器[itBegin,itEnd)的和
// Parameter: const vector<UINT32>::const_iterator & itBegin
// Parameter: const vector<UINT32>::const_iterator & itEnd
// Returns:   UINT32
//************************************
UINT64 GetSum(const vector<UINT32>::const_iterator &itBegin,
        const vector<UINT32>::const_iterator &itEnd)
{
    UINT64 sum = 0;

    for(vector<UINT32>::const_iterator it = itBegin; it != itEnd; ++it)
    {
        sum += *it;
    }

    return sum;
}


void TestFun1()
{


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

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

    // TestFun1();

    DetailPrinter detailPrinter;
    detailPrinter.Start();

    /*********************************算法開始*******************************/
    const UINT32 MAX_NUM = 1000000;
    vector<UINT32> primers;

    MakePrimerVector(MAX_NUM - 1, primers);

    UINT32 count = primers.size();
    UINT64 currSum = 0;
    bool notFound = true;

    vector<UINT32>::const_iterator itBegin; // 數列的第一個數
    vector<UINT32>::const_iterator itEnd;   // 數列的最後一個數的下一個元素！

    itBegin = primers.begin();

    // numCount為數列中的數字個數
    for(UINT32 numCount = count; numCount > 0 && notFound; --numCount)
    {
        itEnd = itBegin + numCount;

        // 擷取第一組數列，索引範圍[0,numCount-1]
        currSum = GetSum(itBegin, itEnd);

        if(currSum < MAX_NUM && MoonMath::IsPrimer(currSum))
        {
            break;
        }

        UINT32 remainSetCount = count - numCount;   // 剩餘的數列個數

        for(UINT32 setIndex = 0; setIndex < remainSetCount; ++setIndex)
        {
            // 計算下一組數列和，減去前一個數，加上後一個數
            currSum = currSum - primers[setIndex] + primers[setIndex + numCount];

            if(currSum < MAX_NUM && MoonMath::IsPrimer(currSum))
            {
                itBegin += setIndex + 1;
                itEnd += setIndex + 1;

                notFound = false;
                break;
            }
        }
    }

    cout << currSum << " is sum of " << *itBegin << " to " << *(itEnd - 1) << endl;

    cout<<"sum is "<<GetSum(itBegin,itEnd)<<endl;

    /*********************************算法結束*******************************/

    detailPrinter.End();
}

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

/*
void F1()
997651 is sum of 7 to 3931
sum is 997651
Total Milliseconds is 1.60107e+006
*/