【ProjectEuler】ProjectEuler_050（找到100萬以內最多連續素數的和，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的範圍內是否很接近整數    //            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，返回最接近的整數值    // Returns:   bool                  接近返回true，否則返回false    //************************************    static bool IsInt(double doubleValue, double epsilon, INT32 &intValue);    //************************************    // Method:    Sign    // Access:    public    // Describe:  擷取value的符號    // Parameter: T value   要擷取符號的值    // Returns:   INT32     正數、0和負數分別返回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          是素數返回true，否則返回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:  計算迭代器[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();    /*********************************演算法開始*******************************/    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;    /*********************************演算法結束*******************************/    detailPrinter.End();}//主函數int main(){    F1();    return 0;}/*void F1()997651 is sum of 7 to 3931sum is 997651Total Milliseconds is 1.60107e+006*/