求除数对

我正在尝试解决这个练习http://main.edu.pl/en/archive/amppz/2014/dzi，我不知道如何提高我的代码的性能。当程序必须处理超过500,000个唯一数字(如描述中最多2,000,000个)时，就会出现问题。然后用1-8秒循环遍历所有这些数字。我使用的测试来自http://main.edu.pl/en/user.phtml?op=tests&c=52014&task=1263，我通过命令
进行测试program.exe < data.in > result.out

描述:You are given a sequence of n integer a1, a2, ... an. You should determine the number of such ordered pairs(i, j), that i, j equeals(1, ..., n), i != j and ai is divisor of aj.
The first line of input contains one integer n(1 <= n <= 2000000) The second line contains a sequence of n integers a1, a2, ..., an(1 <= ai <= 2000000).
In the first and only line of output should contain one integer, denoting the number of pairs sought.
For the input data: 5 2 4 5 2 6
the correct answer is: 6
Explanation: There are 6 pars: (1, 2) = 4/2, (1, 4) = 2/2, (1, 5) = 6/2, (4, 1) = 2/2, (4, 2) = 4/2, (4, 5) = 6/2.

:
-总数为2M，唯一编号为635k，总共有3.45亿次迭代
-总共有2M个数字和200万个唯一数字，总共有1.885亿次迭代

#include <iostream>
#include <math.h>
#include <algorithm>
#include <time.h>

#define COUNT_SAME(count) (count - 1) * count

int main(int argc, char **argv) {
    std::ios_base::sync_with_stdio(0);
    int n; // Total numbers
    scanf("%d", &n);
    clock_t start, finish;
    double  duration;
    int minVal = 2000000;
    long long *countVect = new long long[2000001]; // 1-2,000,000; Here I'm counting duplicates
    unsigned long long counter = 0;
    unsigned long long operations = 0;
    int tmp;
    int duplicates = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &tmp);
        if (countVect[tmp] > 0) { // Not best way, but works
            ++countVect[tmp];
            ++duplicates;
        } else {
            if (minVal > tmp)
                minVal = tmp;
            countVect[tmp] = 1;
        }
    }
    start = clock();
    int valueJ;
    int sqrtValue, valueIJ;
    int j;
    for (int i = 2000000; i > 0; --i) {
        if (countVect[i] > 0) { // Not all fields are setted up
            if (countVect[i] > 1) 
                counter += COUNT_SAME(countVect[i]); // Sum same values
            sqrtValue = sqrt(i);
            for (j = minVal; j <= sqrtValue; ++j) {
                if (i % j == 0) {
                    valueIJ = i / j;
                    if (valueIJ != i && countVect[valueIJ] > 0 && valueIJ > sqrtValue)
                        counter += countVect[i] * countVect[valueIJ];
                    if (i != j && countVect[j] > 0)
                        counter += countVect[i] * countVect[j];
                }
                ++operations;
            }
        }
    }
    finish = clock();
    duration = (double)(finish - start) / CLOCKS_PER_SEC;
    printf("Loops time: %2.3f", duration);
    std::cout << "sn";
    std::cout << "nnCounter: " << counter << "n";
    std::cout << "Total operations: " << operations;
    std::cout << "nDuplicates: " << duplicates << "/" << n;
    return 0;
}  

我知道，我不应该在一开始就对数组进行排序，但是我不知道如何使它更好。

任何提示将是伟大的，谢谢!

这里是改进的算法- 0.5s内2M唯一数。感谢@ pjtrail !

#include <iostream>
#include <math.h>
#include <algorithm>
#include <time.h>

#define COUNT_SAME(count) (count - 1) * count

int main(int argc, char **argv) {
    std::ios_base::sync_with_stdio(0);
    int n; // Total numbers
    scanf("%d", &n);
    clock_t start, finish;
    double  duration;
    int maxVal = 0;
    long long *countVect = new long long[2000001]; // 1-2,000,000; Here I'm counting duplicates
    unsigned long long counter = 0;
    unsigned long long operations = 0;
    int tmp;
    int duplicates = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &tmp);
        if (countVect[tmp] > 0) { // Not best way, but works
            ++countVect[tmp];
            ++duplicates;
        } else {
            if (maxVal < tmp)
                maxVal = tmp;
            countVect[tmp] = 1;
        }
    }
    start = clock();
    int j;
    int jCounter = 1;
    for (int i = 0; i <= maxVal; ++i) {
        if (countVect[i] > 0) { // Not all fields are setted up
            if (countVect[i] > 1)
                counter += COUNT_SAME(countVect[i]); // Sum same values
            j = i * ++jCounter;
            while (j <= maxVal) {
                if (countVect[j] > 0)
                    counter += countVect[i] * countVect[j];
                j = i * ++jCounter;
                ++operations;
            }
            jCounter = 1;
        }
    }
    finish = clock();
    duration = (double)(finish - start) / CLOCKS_PER_SEC;
    printf("Loops time: %2.3f", duration);
    std::cout << "sn";
    std::cout << "nnCounter: " << counter << "n";
    std::cout << "Total operations: " << operations;
    std::cout << "nDuplicates: " << duplicates << "/" << n;
    return 0;
}