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傳送門 Street Numbers

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?2529	?	Accepted:?1406

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):?
6 835 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

6 835 49

Source

New Zealand 1990 Division I,UVA 138 題目要求解1+2+3+...+n=(n+1)+....+m. 要使1+2+3+...+n=(n+1)+....+m.那么n(n+1)/2=(m-n)(n+n+1)/2,即(2m+1)^2-8n^2=1.令x=2m+1，y=n，有x^2-8y^2=1.因此就變成解佩爾方程，而x1=3,y1=1.根據迭代公式： xn=xn-1*x1+d*yn-1*y1 yn=xn-1*y1+yn-1*x1.

已知x1=3，y1=1；

有佩爾方程的迭代公式：x[n]=x[n-1]*x1+d*y[n-1]*y1. ? y[n]=y[n-1]*y1+yn-1]*x1

故有：x[n+1]=3x[n]+8y[n];

? ? ? ? ? ?y[n+1]=x[n]+3y[n];

就可以求出前十組解。 #include<stdio.h> #include<math.h> int main() { int x1=3,y1=1,d=8,x,y,px=3,py=1; for(int i=1;i<=10;i++) { x=x1*px+d*y1*py; y=y1*px+x1*py; printf("%10d%10d\n",y,(x-1)/2); px=x;py=y; } return 0; }

?

Pell方程，由費馬提出，但后來歐拉誤記為佩爾提出，并寫入他的著作中。后人多稱佩爾方程。

設d是正整數，且非平方數。 下面的不定方程稱為佩爾（Pell）方程： (1)一定有無窮多組正整數解 這是初等數論中最經典的內容之一。 推導過程

轉載于:https://www.cnblogs.com/kimsimple/p/7342632.html

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