POJ 1305-Fermat vs. Pythagoras（本原的畢達哥拉斯三元組+列舉）

kewlgrl發表於2016-08-04
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Fermat vs. Pythagoras
Time Limit: 2000MS   Memory Limit: 10000K
Total Submissions: 1555   Accepted: 908

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

題目意思：

給出一個數N，求出所有小於N的X，Y，Z，使之滿足X^2+Y^2=Z^2。
輸出滿足條件的三元組的個數，以及小於N的範圍內三元組不涉及的數的個數。

解題思路：

列舉，且根據定義，本原畢達哥拉斯三元組滿足：
X=m^2-n^2；
Y=2*m*n;
Z=m^2+n^2. 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#define MAXN 1000010
typedef long long ll;
using namespace std;

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
void solve(int t)
{
    int temp=sqrt(t);
    int i,m,n,x,y,z,ans1=0,ans2=0;
    bool flag[MAXN];//標識是否涉及該數
    memset(flag,false,sizeof(flag));
    for(n=1; n<=temp; ++n)//列舉
        for(m=n+1; m<=temp; ++m)
        {
            if(m*m+n*n>t) break;
            if(m%2!=n%2)
                if(gcd(m,n)==1)
                {
                    x=m*m-n*n;//求出三元組
                    y=2*m*n;
                    z=m*m+n*n;
                    ++ans1;
                    //cout<<x<<" "<<y<<" "<<z<<endl;
                    for(i=1;; ++i)//滿足條件的三元組的整數倍也滿足
                    {
                        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;
    cout<<ans1<<" "<<ans2<<endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    while(cin>>n)
    {
        solve(n);
    }
    return 0;
}

/**
10
25
100
**/



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