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): 1749    Accepted Submission(s): 658

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

 

 

Recommend

JGShining

 

题目链接

 

解题思路：

n/1=1/(n+k)+k/(n*(n+k))=1/(n+k)+1/(n^2+n*k)/k=1/(n+k)+1/(n^2/k+n);

 

由于a=n+k<2*n(如果等于2n那么就是a==b了) 得到 k<n，而显然k是n^2的因子

 

方法：质数打表+DFS

 

 

#include<stdio.h>

__int64 pri[1000011];
bool num[1000011];

struct Node{
	__int64 x;
	int cent;
}map[1000];
int d[1000];
__int64 s,n,count;

void Is_prime()
{
	__int64 i,j;
	pri[0]=0;
	for(i=2;i<=1000000;i++)
	{
		if(!num[i])
		{
			pri[++pri[0]]=i;
			for(j=i*i;j<=1000000;j+=i)
				num[j]=1;
		}
	}
}

void DFS(__int64 tmp,int x)
{
	int i;
	for(i=x;i<=count;i++)
	{
		if(map[i].cent>0 && tmp*map[i].x<=n)
		{
			s++;
			map[i].cent--;
			DFS(tmp*map[i].x,i);
			map[i].cent++;
		}
		else if(tmp*map[i].x>n)
			return ;

	}
}

int main()
{
	int T,t=1;
	__int64 i,tmp,m; 

	Is_prime();

	scanf("%d",&T);

	while(t<=T)
	{
		scanf("%I64d",&n);
		printf("Scenario #%d:\n",t);

		count=0;
		tmp=n;
		for(i=1;pri[i]<=tmp && i<=pri[0];i++)
		{
			if(tmp%pri[i]==0)
			{
				map[++count].x=pri[i];
				map[count].cent=0;
				while(!(tmp%pri[i]))
				{
					map[count].cent+=2;
					tmp/=pri[i];
				}
			}
		}
		if(tmp!=1)
		{
			map[++count].x=tmp;
			map[count].cent=2;
		}

		s=0;
		for(i=1;i<=count;i++)
		{
			if(map[i].x<=n)
			{
				s++;
				map[i].cent--;
				DFS(map[i].x,i);
				map[i].cent++;
			}
			else
				break;
		}

		printf("%I64d\n\n",++s);
		t++;
	}
	return 0;
}