【板子】计算一个数的因子的个数

Let’s denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

Find the sum modulo 1073741824 (2^30).
Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).
Output
Print a single integer — the required sum modulo 1073741824 (230).
Examples
Input
2 2 2
Output
20
Input
5 6 7
Output
1520

Note

For the first example.

    d(1·1·1) = d(1) = 1;
    d(1·1·2) = d(2) = 2;
    d(1·2·1) = d(2) = 2;
    d(1·2·2) = d(4) = 3;
    d(2·1·1) = d(2) = 2;
    d(2·1·2) = d(4) = 3;
    d(2·2·1) = d(4) = 3;
    d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.
1
2
3
4
5
6
7
8
9
10
11
12

#include <iostream>
using namespace std;

typedef long long ll;


ll d(int n)//求因子个数--先进行质因数分解，然后再求各个因数的（幂+1）相乘
{
    ll ans = 1;
    for (int i = 2; i * i <= n; ++i)
    {
        if (n % i == 0)
        {
            ll counter=0;
            while(n%i==0)
            {
                n/=i;
                counter++;
            }
            ans = ans*(counter+1);
        }
    }
    if(n>1) ans*=2;//质数的因子有两个
    return ans;
}

int main()
{
    ll a, b, c;
    cin >> a >> b >> c;
    ll ans = 0;
    for (int i = 1; i <= a; ++i)
    {
        for (int j = 1; j <= b; ++j)
            for (int k = 1; k <= c; ++k)
            {
                ans += d(i * j * k);
                //cout << ans << endl;
            }
    }
    cout << ans%1073741824 << endl;
    //system("pause");
}