hdu 1129 Diophantus of Alexandria

Problem Description
Diophantus of Alexandria was an egypt mathematician living in Alexandria. He was one of the first mathematicians to study equations where variables were restricted to integral values. In honor of him, these equations are commonly called diophantine equations. One of the most famous diophantine equation is x^n + y^n = z^n. Fermat suggested that for n > 2, there are no solutions with positive integral values for x, y and z. A proof of this theorem (called Fermat’s last theorem) was found only recently by Andrew Wiles.

Consider the following diophantine equation:

1 / x + 1 / y = 1 / n where x, y, n ∈ N+ (1)

Diophantus is interested in the following question: for a given n, how many distinct solutions (i. e., solutions satisfying x ≤ y) does equation (1) have? For example, for n = 4, there are exactly three distinct solutions:

1 / 5 + 1 / 20 = 1 / 4
1 / 6 + 1 / 12 = 1 / 4
1 / 8 + 1 / 8 = 1 / 4

Clearly, enumerating these solutions can become tedious for bigger values of n. Can you help Diophantus compute the number of distinct solutions for big values of n quickly?

Input
The first line contains the number of scenarios. Each scenario consists of one line containing a single number n (1 ≤ n ≤ 10^9).

Output
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Next, print a single line with the number of distinct solutions of equation (1) for the given value of n. Terminate each scenario with a blank line.

Sample Input

2
4
1260

Sample Output

Scenario #1:
3

Scenario #2:
113

分析：题意就是给定n,求 x/1+y/1=n/1 满足个数。这应该是第2次做这类的题了，这种题解题思路都比较巧妙！由于已知条件少，一般会根据概侓,唯一分解定理等数论知识求出。像这题的话就是：y=k+n 代入 得 x=n^2/k+n,从而转换成求n*n的约数个数，求出x的个数。自己的原解题思路是错了。
补充：N=(p1^a1)(p2^a2)(p3^a3)…(pm^am)
约数个数：(a1+1)(a2+1)(a3+1)…(am+1)
N*N 约数就为：(a1*2+1)(a2*2+1)(a3*2+1)…(am*2+1).

#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
#include<cmath>
#include<queue>
#include<set>
#include<stack>
using namespace std;
const int maxn=50000;
bool vis[100000];
int p[100000],num[100000];
int cnt,len=0;

void is_prime()//素数打表
{
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])
        {
            p[len++]=i; 
            for(int j=i+i;j<maxn;j+=i)
            {
                vis[j]=1;
            }
        }
    }
}

void Dec(int  x)//唯一分解定理求因子幂
{
    memset(num,0,sizeof(num));
    cnt = 0;
    for(int i=0; p[i]*p[i]<=x&&i<len;i++)
    {
        if(x%p[i] == 0)
        {
            while(x%p[i] == 0)
            {

                num[cnt]++;
                x /= p[i];
            }
            cnt++;
        }
    }
    if(x > 1)
    {
        num[cnt++] = 1; 
    }
}

 int main()
 {
    int t,n,sum;
    is_prime();
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        sum=1;
        cin>>n;
        cout<<"Scenario #"<<i<<":"<<endl;
        Dec(n);
        for(int j=0;j<cnt;j++)
        {
        sum*=(num[j]*2+1); 
        }
         cout<<sum/2+1<<endl;
         cout<<endl;
     }
     return 0;
  }