数论(毕达哥拉斯定理)：POJ 1305 Fermat vs. Pythagoras

Fermat vs. Pythagoras

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 1493		Accepted: 865

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(驱使)
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(除数)
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：给定一个n,输出三元组(a,b,c)其中GCD（a，b，c）=1，且a²+b²=c²,以及1~n中没有在任何一个三元组中出现过的数的个数。
　　这道题需要知道勾股数的性质。
　　最最开始GCD（a,b,c）=1 ==> a,b,c两两互质，通过a²+b²=c²易证。
　　首先，对于一组勾股数，①a与b的奇偶性不同，②c一定为奇数。
　　证明①：若a与b同为偶数，则c也为偶数，与GCD（a，b，c）=1矛盾；若a与b同为奇数，c一定为偶数，设a=2*i+1，b=2*j+1，c=2*k -> a²+b²=c²->2*i²+2*i+2*j²+2*j+1=2*k²,这个式子是矛盾的。
　　证明②：a，b一奇一偶，显然。
　　
　　然后将 a²+b²=c² 变形，b为偶数时，a²=(c+b)*(c-b),容易发现c-b与c+b互质，所以c-b与c+b都是平方数，设x²=c-b，y²=c+b，得a=x*y，b=(y²-x²)/2,c=(y²+x²)/2.
　　易得x，y都为奇数,直接枚举x，y就好了。

　　

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int maxn=;
 bool vis[maxn];
 long long Gcd(long long a,long long b){
     return b?Gcd(b,a%b):a;
 }
 int main(){
     int n,m,ans,tot;
     while(~scanf("%d",&n)){
         m=(int)sqrt(n+0.5);ans=tot=;
         memset(vis,,sizeof(vis));
         for(int t=;t<=m;t+=)
             for(int s=t+;(s*s+t*t)/<=n;s+=)
                 if(Gcd(s,t)==){
                     int a=s*t;
                     int b=(s*s-t*t)/;
                     int c=(s*s+t*t)/
                     ans++;
                     for(int k=;k*c<=n;k++){
                         vis[k*a]=true;
                         vis[k*b]=true;
                         vis[k*c]=true;
                     }
                 }
         for(int i=;i<=n;i++)
             if(!vis[i])
                 tot++;
         printf("%d %d\n",ans,tot);
     }
 }