毕达哥拉斯三元组（勾股数组）poj1305

本原毕达哥拉斯三元组是由三个正整数x,y,z组成，且gcd(x,y,z)=1,x*x+y*y=z*z

对于所有的本原毕达哥拉斯三元组（a，b，c） （a*a+b*b=c*c,a与b必定奇偶互异，且c为奇数。这里我们设b为偶数）

则：和

a=st

b=(s*s-t*t)/2

c=(s*s+t*t)/2

其中s>t>=1且gcd(s,t)=1 

是一一对应的。

看看别人得证明：

http://blog.csdn.net/loinus/article/details/7824841

看看我的证明

有了这个定理就这题就很好做了。

Fermat vs. Pythagoras

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 1456		Accepted: 848

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

//
//  main.cpp
//  poj1305
//
//  Created by 陈加寿 on 15/11/30.
//  Copyright (c) 2015年 陈加寿. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;

int mark[1001000];

long long gcd(long long a,long long b)
{
    if(b==0) return a;
    return gcd(b,a%b);
}

int main(int argc, const char * argv[]) {
    int n;
    while(cin>>n)
    {
        memset(mark,0,sizeof(mark));
        int cnt=0;
        for(long long i=1;;i+=2)
        {
            long long a,b,c;
            if( (i*i+(i+2)*(i+2))/2 >n ) break;
                
            for(long long j=i+2;;j+=2)
            {
                if(gcd(i,j)==1)
                {
                    c=(i*i+j*j)/2;
                    b=(j*j-i*i)/2;
                    a=i*j;
                    if(c>n) break;
                    cnt++;
                    for(long long k=1;k*c<=n;k++)
                    {
                        mark[a*k]=1;
                        mark[b*k]=1;
                        mark[c*k]=1;
                    }
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)
            if(mark[i]==0) ans++;
        cout<<cnt<<" "<<ans<<endl;
    }
    return 0;
}