【ProjectEuler】ProjectEuler_021

// Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).// If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.//// For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.//// Evaluate the sum of all the amicable numbers under 10000.#include <iostream>#include <windows.h>using namespace std;const int MAX_NUM = 10000;int number[MAX_NUM + 1];//获取某数亲和数int GetAmicableNum(int num){    //寻找此数的亲和数    int amicableNum = 0;    for(int i = num / 2; i >= 1; i--)    {        if(num % i == 0)        {            amicableNum += i;        }    }    //记录此数的亲和数    number[num] = amicableNum;    //返回亲和数    return amicableNum;}void F1(){    cout << "void F1()" << endl;    LARGE_INTEGER timeStart, timeEnd, freq;    QueryPerformanceFrequency(&freq);    QueryPerformanceCounter(&timeStart);    int result = 0;//亲和数的数目    int tempAmicableNum = 0;    for(int i = 1; i <= MAX_NUM; i++)    {        tempAmicableNum = GetAmicableNum(i);        //是亲和数        if(tempAmicableNum <= MAX_NUM && number[tempAmicableNum] == i && tempAmicableNum != i)        {            cout << number[i] << " " << i << endl;            result += i + tempAmicableNum;        }    }    cout << MAX_NUM << "以内的亲和数之和为" << result << endl;    QueryPerformanceCounter(&timeEnd);    cout << "Total Milliseconds is " << (double)((double)(timeEnd.QuadPart - timeStart.QuadPart) * 1000 / (double)freq.QuadPart) << endl;}//主函数int main(){    F1();    return 0;}/*void F1()220 2841184 12102620 29245020 55646232 636810000以内的亲和数之和为31626Total Milliseconds is 216.438By GodMoon*/