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 10
毕达哥拉斯三元方程组
满足方程x^2+y^2=z2的三元组(x,y,z)(x,y,z)为毕达哥拉斯三元组。当gcd(x,y,z)=1，称其为本原。
存在互质的正整数，一奇一偶，m,n满足
x=m^2-n^2
y=2*m*n
z=m^2+n^2
本题就利用了上述结论，找出方程本原，再成倍标记
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define maxn 1000005
using namespace std;
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
bool flag[maxn];
int main()
{
int N;
while(~scanf("%d",&N))
{
int temp,m,n,ans1,ans2,x,y,z;
ans1=ans2=0;
memset(flag,false,sizeof(flag));
temp=sqrt(N+0.5);
for(int n=1;n<=temp;++n)
{
for(int m=n+1;m<=temp;++m)
{
if(m*m+n*n>N)
break;
if((n&1)!=(m&1))
{
if(gcd(m,n)==1)
{
x=m*m-n*n;
y=2*m*n;
z=m*m+n*n;
ans1++;
int i;
for( i=1;;++i)
{
if(i*z>N)
break;
flag[i*x]=true;
flag[i*y]=true;
flag[i*z]=true;
}
}
}
}
}
for(int i=1;i<=N;i++)
if(!flag[i])
ans2++;
printf("%d %d",ans1,ans2);
printf("\n");
}
return 0;
}