UVa 106 - Fermat vs Pythagoras（数论题目）

题目来源：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=42

Fermat vs. Pythagoras

Background

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 for n > 2.

The Problem

Given a positive integer N, you are to write a program that computes two quantities regarding the solution of

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 such that p is not part of any triple (not just relatively prime triples).

The 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.

The 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 ). The second number is the number of positive integers that are not part of any triple whose components are all . There should be one output line for each input line.

Sample Input

Sample Output

1 4

4 9

16 27
解题思路：

这是一道数论题，用数学的语言描述就是：x, y, z∈N，给定一个数n，找出所有的x, y, z ≤ n，使得x² + y² = z²成立。如果要穷举所有的x, y, z的话，按照题目所给的数据量，肯定是无法在限定时间内完成的。考虑利用毕达哥拉斯数的性质生成所有的x, y, z来解决，数学推导简要介绍如下：

先假定x, y, z两两互质，由于x, y互质，故x, y中至少有1个是奇数。下面用反证法证明x和y中有且只有1个奇数。假定x, y都为奇数，设：

x = 2a + 1
y = 2b + 1
x² + y² = (2a + 1)² + (2b + 1)²
= 4(a² + b² + a + b) + 2

又因为x²和y²是奇数，则z²是偶数，且必能被4整除，与上式矛盾，因此x, y中只有一个奇数。

假设x为奇数，y为偶数，则z为奇数，2z与2x的最大公因数为2，2z和2x可分别写作

2z = (z + x) + (z - x)
2x = (z + x) - (z - x)

那么跟据最大公因数性质，z + x和z - x的最大公因数也为2，又因为：

(z + x)(z - x) = y²，两边同除以4得：
((z + x) / 2)((z - x) / 2) = (y / 2)²

故可令：

z + x = 2m², z - x = 2n²
其中z = m + n, x = m - n（m与n互质）

则有：

y² = z² - x² = 2m²2n² = 4m²n²
即y = 2mn。

综上所述，可得到下式：

x = m² - n², y = 2mn, z = m² + n². (m, n为任意自然数)

这里还有一个问题：题目要求统计(x, y, z)三元组的数量时只统计x，y和z两两互质的的情况，这个问题用上面的算法就可以解决了。但对于统计p的数量，题目并不限定三元组是两两互质的。但是上式不能生成所有x, y, z并不是两两互质的情况。然而假设x与y最大公因数w不为1，则z也必能被w整除，因此w为x, y, z三个数的公因数。归纳总结可知，所有非两两互质的x₀, y₀, z₀都可由一组互质的x, y, z乘以系数得到。根据以上理论就可以快速的求解了。


参考代码：

 #include <cstdio>

 #include <cmath>

 #include <cstring>

 #define N 1000010

 bool used[N];

 long long gcd(long long a , long long b)

 { return b== ? a: gcd(b,a%b); }

 int main()

 {

     long long n,a,b,c;

     long long count1,count2;

     while(scanf("%lld",&n)!=EOF)

     {

         count1=count2=;

         memset(used,,sizeof(used));

         long long m=(long long)sqrt(n+0.5);

         for(long long t=; t<=m; t+=)

             for(long long s=t+; s*t<=n; s+=)

                 if(gcd(s,t)==)  //s>t>=1且s与t互质

                 {

                     a=s*t;          //奇数

                     b=(s*s-t*t)/;  //偶数

                     c=(s*s+t*t)/;  //奇数

                     if(c<=n)        //在n范围内的PPT

                     {

                         count1++;

                         //printf("本原勾股数组：%lld %lld %lld\n",a,b,c);

                         if(!used[a]) { count2++; used[a]=; }

                         if(!used[b]) { count2++; used[b]=; }

                         if(!used[c]) { count2++; used[c]=; }

                         for(int j=; c*j<=n; j++)  //j是倍数

                         {

                             if(!used[a*j]) { count2++; used[a*j]=; }

                             if(!used[b*j]) { count2++; used[b*j]=; }

                             if(!used[c*j]) { count2++; used[c*j]=; }

                         }

                     }

                 }

             printf("%lld %lld\n",count1,n-count2);

     }

     return ;

 }