【数论】HDU 1299 Diophantus of Alexandria 【任一分解定理+素数打表】

最新推荐文章于 2019-04-09 20:42:39 发布

原创最新推荐文章于 2019-04-09 20:42:39 发布 · 304 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

18 篇文章

订阅专栏

本文探讨了Diophantus方程1/x + 1/y = 1/n的求解方法，特别是对于给定的大整数n，如何快速计算方程的不同解的数量。文章介绍了一种有效的算法实现，并通过样例输入输出展示了该方法的有效性。

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

#include<bits/stdc++.h>
using namespace std;
int prime[1000005];
void makeprime()
{
    int  total=0;
    int flag;
    prime[total++]=2;
    for(int i=3; i<=1000000; i++)
    {
        flag=1;
        for(int j=0; prime[j]*prime[j]<=i; j++)
        {
            if(i%prime[j]==0)
            {
                flag=0;
                break;
            }
        }
        if(flag)
            prime[total++]=i;
    }
}
int main()
{
    makeprime();
    int t,n;
    cin>>t;
    int cas=1;
    while(t--)
    {
        cin>>n;
        int ans=1;
        for(int i=0; prime[i]<=sqrt(n); i++)
        {
            int sum=0;
            while(n%prime[i]==0)
            {
                sum++;
                n/=prime[i];
            }
            ans*=(2*sum+1);
        }
        if(n>1)
            ans*=3;
        printf("Scenario #%d:\n",cas++);
        printf("%d\n\n",(ans+1)/2);

    }

}