HDU 1299 Diophantus of Alexandria (公式變形 分解質因數)

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公式


Diophantus of Alexandria

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2931    Accepted Submission(s): 1132

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




 

Source 

TUD Programming Contest 2006


題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1299


題目大意：給出n，求滿足1/x + 1/y = 1/n的(x,y)的對數，(x,y)與(y,x)屬於同一種


題目分析：1/x + 1/y = 1/n => xy - nx - ny = 0 => n^2 + xy - nx - ny = n^2 => (n - x)(n - y) = n^2，問題轉化為求n^2的因子數，由唯一分解定理可以得到：設p1,p2...pk為n的質因子a1,a2...ak，為質因子的冪，則n^2因子數為(2*a1 + 1)*(2*a2 + 1)*...*(2*ak + 1)，注意當n為質數時根據公式要乘3，考慮去掉重複的最後答案除2加1
#include <cstdio>
#define ll long long
int const MAX = 4e5 + 5;
int p[MAX], pfac[MAX];
bool noprime[MAX];
int pnum, pfacnum, n;
ll ans;

void get_prime()
{
    pnum = 0;
    for(int i = 2; i < MAX; i++)
    {
        if(!noprime[i])
            p[pnum ++] = i;
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)
        {
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)
                break;
        }
    }
}

void get_pfac()
{
    pfacnum = 0;
    for(int i = 2; i * i <= n; i++)
    {
        int cnt = 0;
        if(n % i == 0)
        {
            pfac[pfacnum ++] = i;
            while(n % i == 0)
            {
                cnt ++;
                n /= i;
            }
        }
        if(cnt)
            ans *= (ll) (2 * cnt + 1);
    }
}

int main()
{
    get_prime();
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        ans = 1;
        scanf("%d", &n);
        get_pfac();
        if(n > 1)
            ans *= 3ll;
        printf("Scenario #%d:\n%I64d\n\n", ca, (ans + 1) >> 1);
    }
}




 





            




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