【ProjectEuler】ProjectEuler_037

// Problem 37// 14 February 2003//// The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.//// Find the sum of the only eleven primes that are both truncatable from left to right and right to left.//// NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.#include <iostream>#include <windows.h>#include <cmath>#include <ctime>using namespace std;// 判斷某數是否為素數bool IsPrimeNum(int num){    if((num % 2 == 0 && num > 2) || num <= 1)    {        return false;    }    int sqrtNum = (int)sqrt((double)num);    for(int i = 3; i <= sqrtNum; i += 2)    {        if(num % i == 0)        {            return false;        }    }    return true;}// 判斷是否為Truncatable素數bool CheckTruncatableNum(const int num){    int currentNum = num;    //從右往左剔除數字，此處已經判斷過原始數了，下面就不用判斷了    while(currentNum != 0)    {        if(!IsPrimeNum(currentNum))        {            return false;        }        currentNum /= 10;    }    //從左往右剔除數字    int tenDigit = 10;    currentNum = num % tenDigit;    while(currentNum != num)    {        if(!IsPrimeNum(currentNum))        {            return false;        }        tenDigit *= 10;        currentNum = num % tenDigit;    }    return true;}void F1(){    cout << "void F1()" << endl;    LARGE_INTEGER timeStart, timeEnd, freq;    QueryPerformanceFrequency(&freq);    QueryPerformanceCounter(&timeStart);    const int MIN_NUM = 11;//從11開始，因為題目要求排除2,3,5,7    const int MAX_COUNT = 11;//總共有11個    int sum = 0;//記錄總和    int count = 0;//記錄總數    for(int i = MIN_NUM; count < MAX_COUNT; i += 2)    {        if(CheckTruncatableNum(i))        {            cout << i << endl;            count++;            sum += i;        }    }    cout << "總和為" << sum << endl;    QueryPerformanceCounter(&timeEnd);    cout << "Total Milliseconds is " << (double)(timeEnd.QuadPart - timeStart.QuadPart) * 1000 / freq.QuadPart << endl;    time_t currentTime = time(NULL);    char timeStr[30];    ctime_s(timeStr, 30, ¤tTime);    cout << endl << "By GodMoon" << endl << timeStr;}//主函數int main(){    F1();    return 0;}/*void F1()2337537331331737379731373797739397總和為748317Total Milliseconds is 453.591By GodMoonSat Nov 05 14:09:20 2011*/