POJ 1320 Street Numbers 【佩尔方程】

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博客围绕 POJ 1320 题目展开，题意为判断 M 个编号 1 到 M 的房子，是否存在 1 到 N - 1 的和等于 N + 1 到 M 的和，并求解前十个 N 和 M。解题思路是通过两个等差数列求和化简得到（2M + 1）^2 - 8N^2 = 1，再利用佩尔方程递推式求解。

任意门：http://poj.org/problem?id=1320

Street Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3181		Accepted: 1776

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

Source

New Zealand 1990 Division I,UVA 138

题意概括：

有 M 个房子，编号为 1~M，是否存在 1+2+3+...+ (N-1) == (N+1)+(N+2)+...+(M)；

求解前十个 N， M；

解题思路：

两个等差数列求和，化简可得：

（2M+1)^2 - 8N^2 = 1；

令 X = 2M+1， Y = N ;

特解： X0 = 3, Y0 = 1;

根据佩尔方程的递推式：

X[ n+1 ] = X[ n ] * X0 + D * Y[ n ] * Y0;

Y[ n+1 ] = X[ n ] * Y0 + X0 * Y[ n ];