HDU 1299 Diophantus of Alexandria

Diophantus of Alexandria

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2360    Accepted Submission(s): 899

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

 题目描述：给出n，求满等式1/n=1/x+1/y的整数对x，y的对数。
由于x和y一定大于n，设y=n+k，带入等式得x=n*n/k+n；由于x，y均为正整数，所以n*n/k是整数，题目变成了求n*n因子的个数；
对于任意的一个整数n，有唯一的分解式n=p1^a1*p2^a2....*pn^an，p1,p2....pn都是素数，由乘法原理，n的因子数为(a1+1)(a2+1)...(an+1)；可推出n*n的因子总数为(2*a1+1)(2*a2+1)...(2*an+1).

#include<stdio.h>#include<string.h>#define N 40000bool notprime[N];int prime[N],p[N];void Chooseprime(){    prime[0]=0;    memset(notprime,false,sizeof(notprime));    notprime[0]=notprime[1]=true;    int i,j;    for(i=2;i<N;i++)        if(!notprime[i])            for(j=2*i;j<N;j+=i)                notprime[j]=true;    for(i=0;i<N;i++)        if(!notprime[i])            prime[++prime[0]]=i;}int main(){    Chooseprime();    int T,n,k=0;    scanf("%d",&T);    while(T--)    {        int t,sum=1;        scanf("%d",&n);        for(int i=1,t=0;i<=prime[0]&&n>=prime[i];i++)        {            t=0;            while(n%prime[i]==0)            {                n/=prime[i];                t++;            }            sum*=2*t+1;        }        if(n>1)sum*=3;//如果n大于1,表明n还是一个素数，n*n会有三个因子        printf("Scenario #%d:\n%d\n\n",++k,(sum+1)/2);//求出来的对数是x，y对数的两倍    }    return 0;}