本原勾股数 POJ_1305

Fermat vs. Pythagoras

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 1105		Accepted: 641

Description

Computer generated and assisted proofs and verification occupy a small niche in the realm of Computer Science. The first proof of the four-color problem was completed with the assistance of a computer program and current efforts in verification have succeeded in verifying the translation of high-level code down to the chip level. 
This problem deals with computing quantities relating to part of Fermat's Last Theorem: that there are no integer solutions of a^n + b^n = c^n for n > 2. 
Given a positive integer N, you are to write a program that computes two quantities regarding the solution of x^2 + y^2 = z^2, where x, y, and z are constrained to be positive integers less than or equal to N. You are to compute the number of triples (x,y,z) such that x < y < z, and they are relatively prime, i.e., have no common divisor larger than 1. You are also to compute the number of values 0 < p <= N such that p is not part of any triple (not just relatively prime triples). 

Input

The input consists of a sequence of positive integers, one per line. Each integer in the input file will be less than or equal to 1,000,000. Input is terminated by end-of-file

Output

For each integer N in the input file print two integers separated by a space. The first integer is the number of relatively prime triples (such that each component of the triple is <=N). The second number is the number of positive integers <=N that are not part of any triple whose components are all <=N. There should be one output line for each input line.

Sample Input

10
25
100

Sample Output

1 4
4 9
16 27

本原勾股数：
概念：一个三元组(a,b,c)，其中a,b,c没有公因数而且满足：a^2+b^2=c^2
首先，这种本原勾股数的个数是无限的，而且构造的条件满足：
a=s*t,b=(s^2-t^2)/2,c=(s^2+t^2)/2
其中s>t>=1是任意没有公因数的奇数！
由以上概念就可以导出任意一个本原勾股数组。

#include "stdio.h"
#include <math.h>

	int n;
	int s,t;
	int temp ;
	int a , b ,c;
	int count,countP;
	bool flag[1000001];

int gcd(int s ,int t)
{
	if(t==0)
		return s;
	else
		gcd(t,s%t);
}
int main()
{
	
	while(scanf("%d",&n)!=EOF)
	{
		temp = sqrt(n);
		count = 0;
		countP = 0 ;
		int i ,j;
		for(s = 1 ; s <=n ; s++)
			flag[s]=true;
		for(s = 3 ; s<=temp*2; s+=2)
		{
			for(t = 1 ; t<s ; t+=2)
			{
				if(gcd(s,t)!=1)
					continue;
				c = (s*s+t*t)/2;
				if(c>n)
					break;
				a = (s*s-t*t)/2;
				b = s*t;
				for(i = 1 ; i<=n/c;i++)
				{
					flag[a*i]=false;
					flag[b*i]=false;
					flag[c*i]=false;
				}
				count++;	
			}
		}
		for(s = 1 ; s<=n ; s++)
			if(flag[s]==1)
				countP++;
		printf("%d %d\n" , count , countP);
	}
}