POJ - 1305 Fermat vs. Pythagoras —— 毕达哥拉斯三元组

Fermat vs. Pythagoras

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 1651		Accepted: 965

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

题意：给定数n，问能找到多少对三元组以及不能形成三元组的数有多少

思路：数据不大，直接按毕达哥拉斯三元组的判定条件暴力跑，(logn)^2的复杂度

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <vector>
#define max_ 1000100
#define inf 0x3f3f3f3f
#define ll long long
#define les 1e-8
#define mod 9901
using namespace std;
int n;
bool vis[1000010];
int main(int argc, char const *argv[]) {
	while(scanf("%d",&n)!=EOF)
	{
		memset(vis,false,sizeof vis);
		int x,y,z;
		int ans=0;
		for(int i=1;i*i<=n;i++)
		{
			for(int j=i+1;j*j<=n;j++)
			{
				if(i*i+j*j>n)
				break;
				if((i&1)!=(j&1)&&__gcd(i,j)==1)
				{
					x=2*i*j;
					y=j*j-i*i;
					z=i*i+j*j;
					ans++;
					int k=1;
					while(k*z<=n)
					{
						vis[k*x]=vis[k*y]=vis[k*z]=true;
						k++;
					}
				}
			}
		}
		int k=0;
		printf("%d ",ans);
		for(int i=1;i<=n;i++)
		{
			if(vis[i]==false)
			{
				k++;
			}
		}
		printf("%d\n",k);
	}
	return 0;
}