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Fermat vs. Pythagoras
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 Source |
题意是说，能构造多少本元勾股数和勾股数，要求构造的数<=n
所谓本元勾股数，就是三个勾股数没有公因数，两两互质。
由本元勾股数扩大k倍，就可以得到其他勾股数。
而构造本元勾股数的方法如下：
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！
引用一段构造正确性的证明：
本原勾股数组（PPT)是一个三元组（a，b，c),其中a，b，c无公因数，且满足a² +b² =c²。
很明显存在无穷多个勾股数组（abc同乘以n），下面研究abc没有公因数的情况，先写出一些本原勾股数组：
case:(3,4,5) (5,12,13) (8,15,17) (7,24,25) (20,21,29)(9,40,41)(12,35,37)(11,60,61)(28,45,53) (33,56,65) (16,63,65)
观察可以看出a，b奇偶性不同且c总是奇数。（用一点技巧可以证明这是正确的）
3² = 5² - 4² = (5-4)(5+4) = 1 × 9
15² = 17²-8² = (17-8)(17+8) = 9 ×25
35² = 37² - 12² = (37-12)(37+12) = 25 ×49
......
很神奇的是似乎c-b与c+b总是平方数，并且c-b与c+b木有公因数。证明一下下：假设有公因数，设d是c-b与c+b的公因数，则d也整除(c+b)+(c-b)=2c, (c+b)-(c-b) = 2b,所以d整除2c，2b，但是b，c木有公因数，又假设了（a，b，c)是本原勾股数组，从而d等于1或2，又因为d整除（c-b)(c+b)=a².a²是奇数，所以d = 1，c-b与c+b木有公因数。，又因为（c-b)(c+b)=a²,所以c-b与c+b的积是平方数，只有二者都是平方数才会出现（可以把二者分解成素数乘积直观地看出），令c+b = s²，c-b=t²，解得
c=（s²+t²)/2, b=(s²-t²)/2,a = √(c-b)(c+b) = st.这就得出了勾股数组定理：
每个本原勾股数组（a，b，c）(a为奇数，b偶数）都可由如下公式得出：a=st，b=（s²-t²）/2, c = (s²+t²)/2, 其中s>t>=1是没有公因数的奇数。
当取t=1时就可以得到上面的许多例子。
/************************************************************************* > File Name: code/poj/1305.cpp > Author: 111qqz > Email: rkz2013@126.com > Created Time: 2015年08月22日 星期六 13时49分30秒 ************************************************************************/ #include<iostream> #include<iomanip> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<map> #include<set> #include<queue> #include<vector> #include<stack> #define y0 abc111qqz #define y1 hust111qqz #define yn hez111qqz #define j1 cute111qqz #define tm crazy111qqz #define lr dying111qqz using namespace std; #define REP(i, n) for (int i=0;i<int(n);++i) typedef long long LL; typedef unsigned long long ULL; const int inf = 0x3f3f3f3f; const int N= 1E6+7; bool v[N]; int n; int gcd(int a,int b){ if (a<b) return gcd(b,a); if (a%b==0) return b; return gcd(b,a%b); } int main() { while (scanf("%d",&n)!=EOF){ memset(v,false,sizeof(v)); int ans = 0 ; int cnt = 0 ; for ( int t = 1 ; t <= n ;t = t + 2){ for ( int s = t+2 ; s*t<= n; s = s + 2){ if (gcd(s,t)==1){ int a = s*t; int b = (s*s-t*t)/2; int c = (s*s+t*t)/2; if (a<=n&&b<=n&&c<=n){ ans++; for ( int i = 1 ; i*a<=n&&i*b<=n&&i*c<=n;i++){ v[i*a] = true; v[i*b] = true; v[i*c] = true; } } } } } for ( int i = 1 ; i <= n ; i++){ if (!v[i]) cnt++; } printf("%d %d ",ans,cnt); } return 0; }