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

 

Sample Output

 

Source

TUD Programming Contest 2006

 

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这题数论求的是 数的因子个数，设数为n，其可表示为

n=p1^r1 * p2^r2 * . . . * pn^rn

其中，p为素数，且可知其n的因子个数

k=(r1+1)*(r2+2)*...*(rn+1);

 

又由题可得，

1/x+y/1=1/n  ==> 

x>n && y>n   ==> 

xy=nx+ny，设y=n+k，x=n*(n+k)/k，即所求为n*n的因子个数

k=(2*r1+1)*(2*r2+2)*...*(2*rn+1);

 

注意结果要求多少对，故ans=k/2+1;

 1 //140MS    200K    622 B    G++ 2 #include<stdio.h> 3 #include<math.h> 4 long long solve(int n) 5 { 6     long long ans=1; 7     int i; 8     int m=(int)sqrt(n+0.5); 9     for(i=2;i<=m;i++){10         int ret=1; 11         if(n%i==0){12             n/=i;13             while(n%i==0){14                 n/=i;ret++;15             }16             ans*=(2*ret+1);17         }18         if(n<i) break;19     }20     if(n>1) ans*=3;21     return ans;22 }23 int main(void)24 {25     int n;26     int cas=1,t;27     scanf("%d",&t);28     while(t--)29     { 30         scanf("%d",&n);31         printf("Scenario #%d:\n%lld\n",cas++,solve(n)/2+1);32         printf("\n");33     }34     return 0;35 }