HDU 1299（数论，求n的素因子个数）

Diophantus of Alexandria

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 459    Accepted Submission(s): 144

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

241260

 

 

Sample Output

Scenario #1:3Scenario #2:113
/**************************************************  首先我们知道x、y都是大于n的数  假设y=n+k （k>=1）    带入1/x+1/y=1/n，求出x=n*(n+k)/k  ->   x=n^2/k + n  x固然是一个整数，所以我们只要知道n*n/k为整数的k的个数(k绝对小于n的)  即n*n的因子数。。。。  每一个数都能分解成n=p1^e1*p2^e2*p3^e3...pr^er   p1....pr是小于n的素数  因子数就是num=(1+e1)*(1+e2)*(1+e3)*....*(1+er)  于是n*n的因子数就是cnt=(1+2*e1)*(1+2*e2)*...*(1+2*er)***************************************************/#include <iostream>#include <math.h>using namespace std;#define N 40000int prime[N];int elem[N];int elem_num;void init()      //筛选法找出sqrt(n)之前的素数，存在elem数组里面{int i,j;for (i=2;i<=N;i++){for (j=2;i*j<N;j++){prime[i*j]=0;}}elem_num=0;for (i=2;i<N;i++){if(prime[i])elem[elem_num++]=i;}}int main(){int t,n,num,cnt,i,flag,ct;memset(prime,1,sizeof(prime));init();while (scanf("%d",&t)!=EOF){ct=0;while (t--){scanf("%d",&n);cnt=1;for(i=0;i<elem_num;i++){flag=(int)sqrt(n*1.0)+1;if(elem[i]>flag)break;num=0;while (n%elem[i]==0){num++;n/=elem[i];}cnt*=(2*num+1);}if(n>1) cnt*=3;  //这步要注意，最后有可能剩下的n本身就是原来n的素因子  printf("Scenario #%d:/n",++ct);printf("%d/n/n",(cnt+1)/2); /************************************************************************//* 看测试数据就知道为什么要(cnt+1)/2了，两两配对，有一对一定是相同的    *//************************************************************************/}}return 0;}