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;
}