POJ 1320：Street Numbers 佩爾方程

Street Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2753		Accepted: 1530

Description

A computer programmer lives in a street with houses numbered consecutively (from 1) down one side of the street. Every evening she walks her dog by leaving her house and randomly turning left or right and walking to the end of the street and back. One night she adds up the street numbers of the houses she passes (excluding her own). The next time she walks the other way she repeats this and finds, to her astonishment, that the two sums are the same. Although this is determined in part by her house number and in part by the number of houses in the street, she nevertheless feels that this is a desirable property for her house to have and decides that all her subsequent houses should exhibit it. 
Write a program to find pairs of numbers that satisfy this condition. To start your list the first two pairs are: (house number, last number): 

         6         8

        35        49

Input

There is no input for this program.

Output

Output will consist of 10 lines each containing a pair of numbers, in increasing order with the last number, each printed right justified in a field of width 10 (as shown above).

Sample Input

Sample Output

         6         8
        35        49

題意是一個計算機程序員住在一個門牌號(從1開始計)連續的街道上，每天晚上她都在這個街道上從頭走到尾，有一天晚上她把她走過的門牌號相加起來，下一次她走另一條路還是相加門牌號，令她吃驚的是(我都不知道這有什麼好吃驚的)，兩個和是相等的。讓打出表house number,last number

題意的意思翻譯過來就是1+2+...+x = x+(x+1)+(x+2)...+y

求x，y。

方程就變爲x*(x+1)/2 = (x+y)(y-x+1)/2 => y^2+y-2*x^2=0 => (2*y+1)^2-8*x^2=0

做這道題的收穫就是通過這道題了解了佩爾方程，它是一個解x^2-d*y^2=1這類方程的方法。

佩爾方程的意思就是x^2-d*y^2=1的第一個解x0，y0已知的話，其餘的值有一個遞推公式了：

X(n)=X(n-1)*x0+d*Y(n-1)*y0

Y(n)=X(n-1)*y0+Y(n-1)*x0

知道了這個之後，程序就好寫了。以後記住解x^2-d*y^2=1的方程有一個簡便算法~

代碼：

#include<iostream>
#include<iomanip>
#pragma warning(disable:4996) 
using namespace std;

int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	long long i,x0=3,y0=1,last_x=3,last_y=1,x,y;
	for(i=1;i<=10;i++)
	{
		x=last_x*x0+8*last_y*y0;
		y=last_x*y0+last_y*x0;

		cout<<setw(10)<<y<<setw(10)<<(x-1)/2<<endl; 

		last_x=x;
		last_y=y;
	}      
	system("pause");
	return 0;
}