HDU 1299 基础数论 分解

给一个数n问有多少种x,y的组合使$\frac{1}{x}+\frac{1}{y}=\frac{1}{n},x<=y$满足，设y = k + n,代入得到$x = \frac{n^2}{k} + n$，也就是求n^2的因子数量 

/** @Date    : 2017-09-08 10:45:12
  * @FileName: HDU 1299 数论 分解.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int pri[N];
int vis[N];
int c = 0;
void prime()
{
	MMF(vis);
	for(int i = 2; i < N; i++)
	{
		if(!vis[i])
			vis[i] = 1, pri[c++] = i;
		for(int j = 0; j < c && i * pri[j] < N; j++)
		{
			vis[i*pri[j]] = 1;
			if(i % pri[j] == 0)
				break;
		}
 	}
}


int main()
{
	prime();
	int T;
	cin >> T;
	int icase = 0;
	while(T--)
	{
		LL n;
		scanf("%lld", &n);
		LL t = n * n;//直接对n^2分解不对？
		LL cnt = 1;
		for(int i = 0; i < c && pri[i] * pri[i] <= n; i++)
		{
			if(n % pri[i] == 0)
			{
				LL tmp = 0;
				while(n % pri[i] == 0 && n)
					n /= pri[i], tmp++;
				cnt *= tmp*2+1;
			}
		}
		if(n > 1)
			cnt *= 3;
		cnt = (cnt + 1) / 2;
		printf("Scenario #%d:\n", ++icase);
		printf("%lld\n\n", cnt);
	}
    return 0;
}