Fermat vs. Pythagoras
Time Limit: 2000MS Memory Limit: 10000K
Total Submissions: 1739 Accepted: 1021
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
Duke Internet Programming Contest 1991,UVA 106
问题链接:POJ1305 HDU1615 UVA106 Fermat vs. Pythagoras
问题简述:(略)
问题分析:
毕达哥拉斯三角形问题。设有不定方程:x^2+y^2=z^2,若正整数三元组(x,y,z)满足上述方程,则称为毕达哥拉斯三元组。若gcd(x,y,z)=1,则称为本原毕达哥拉斯三元组。
有关定理:正整数x,y,z构成一个本原的毕达哥拉斯三元组且y为偶数,当且仅当存在互素的正整数m和n(m>n),其中m和n的奇偶性不同,并且满足x=m^2-n^2,y=2mn,z=m^2+n^2。
这个题计算在n范围内(x,y,z<=n)本原毕达哥拉斯三元组的个数,以及n以内且毕达哥拉斯三元组不涉及数的个数。
程序说明:(略)
参考链接:(略)
题记:(略)
AC的C++程序如下:
/* POJ1305 HDU1615 UVA106 Fermat vs. Pythagoras */
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int N = 1e6;
bool flag[N + 1];
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
int main()
{
int n;
while(~scanf("%d", &n)) {
memset(flag, false, sizeof(flag));
int cnt1 = 0, cnt2 = 0, end = (int)sqrt(n);
for(int a = 1 ; a <= end ; a++ )
for( int b = a + 1; a * a + b * b <= n ; b++) {
if(gcd(a, b) == 1 && a % 2 != b % 2) {
cnt1++;
int x = 2 * a * b;
int y = b * b - a * a;
int z = b * b + a * a;
for(int k = 1 ; k * z <= n ; k++)
flag[k * x] = flag[k * y] = flag[k * z] = true;
}
}
for(int i = 1 ; i <= n ; i++)
if( !flag[i] )
cnt2++;
printf("%d %d\n", cnt1, cnt2);
}
return 0;
}