poj1305毕达哥拉斯不定方程

Fermat vs. Pythagoras

Time Limit: 2000MS Memory Limit: 10000KTotal Submissions: 1140 Accepted: 662

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

1025100

Sample Output

1 44 916 27毕达哥拉斯三元组（本原）满足x=m*m-n*n;y=2*m*n;z=m*m+n*n;其中 m>n,m为奇数，n为偶数或者m为偶数，n为奇数；那么要求所给范围内的本原毕达哥拉斯三元组数，只需对m，n分别枚举即可，然后将三元组乘以i（保证i*z在所给范围内），就可以求出所有的毕达哥拉斯三元组(教材112，数论)
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn=1000001;bool flag[maxn];int gcd(int a,int b){    if(!b)  return a;    else    return gcd(b,a%b);}void solve(int t){    int x,y,z,n,m,ans1,ans2,i,temp;    ans1=ans2=0;    memset(flag,0,sizeof(flag));    temp=sqrt(t+0.0);    for(n=1;n<=temp;n++)    {        for(m=n+1;m<=temp;m++)        {            if(m*m+n*n>t)                break;            if(n%2!=m%2)            {                if(gcd(m,n)==1)                {                    x=m*m-n*n;                    y=2*m*n;                    z=m*m+n*n;                    ans1++;                    for(i=1;;i++)                    {                        if(i*z>t)                            break;                        flag[i*x]=true;                        flag[i*y]=true;                        flag[i*z]=true;                    }                }            }        }    }    for(i=1;i<=t;i++)    {        if(!flag[i])            ans2++;    }    printf("%d %d\n",ans1,ans2);}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        solve(n);    }    return 0;}