Hdu 1299

Diophantus of Alexandria

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2326    Accepted Submission(s): 887

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

令y = n + k (k >= 1)，　则x = n^2/k + n , x为整数，　所以k为n^2的约数，因为x >= y, 所以k <= n, 所以

可以将问题简化为求n^2的不大于ｎ的约数的个数，然后素数分解。

Accepted Code:

 1 /*************************************************************************  2  > File Name: 1299.cpp  3  > Author: Stomach_ache  4  > Mail: sudaweitong@gmail.com  5  > Created Time: 2014年07月10日 星期四 16时39分38秒  6  > Propose:  7  ************************************************************************/

 8 

 9 #include <cmath>

10 #include <string>

11 #include <cstdio>

12 #include <fstream>

13 #include <cstring>

14 #include <iostream>

15 #include <algorithm>

16 using namespace std; 17 

18 int n; 19 int prime[100000]; 20 bool vis[100000]; 21 int cnt = 0; 22 

23 void isPrime() { 24       cnt = 0; 25     int k = (int)sqrt(1000000000.0) + 1; 26       for (int i = 0; i <= k; i++) vis[i] = true; 27     for (int i = 2; i <= k; i++) { 28           if (vis[i]) { 29             prime[cnt++] = i; 30             for (int j = i*i; j <= k; j += i) vis[j] = false; 31  } 32  } 33 } 34 

35 int main(void) { 36  isPrime(); 37     int c = 1; 38     int t; 39     scanf("%d", &t); 40       while (t--) { 41           scanf("%d", &n); 42           int ans = 1; 43         for (int i = 0; i < cnt && prime[i] <= n; i++) { 44               int tmp = 0; 45               while (n % prime[i] == 0) { 46                   tmp++; 47                 n /= prime[i]; 48  } 49             ans *= 1 + 2 * tmp; 50  } 51         if (n > 1) ans *= 3; 52         printf("Scenario #%d:\n", c++); 53         printf("%d\n\n", (ans + 1) / 2); 54  } 55 

56     return 0; 57 }