poj 1320

时间：2015-11-05 00:29:27 阅读：160 评论：0 收藏：0 [点我收藏+]

标签：

Street Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2806		Accepted: 1565

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

先介绍一下什么事佩尔方程

定义：形如x^2 - dy^2 = 1（d>1且d不为完全平方数）的不定方程称为佩尔方程

若佩尔方程x^2 - dy^2 = 1的最小特解是（x1,y1）那么可以有迭代公式

xn = xn-1*x1 + d*yn-1*y1;

yn = xn-1*y1 + yn-1*x1;

上面题目将等式两边化简得到(2*m+1)^2 - 8n^2 = 1然后套用迭代公式就好了

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <algorithm>
 8 #include <vector>
 9 
10 using namespace std;
11 
12 const int maxn = 1000005;
13 
14 int main()
15 {
16     int x,y,x1,y1,px,py,d;
17     x1 = 3;
18     y1 = 1;
19     px = 3;
20     py = 1;
21     d = 8;
22     for(int i=1;i<=10;i++){
23         x = px*x1 + d*py*y1;
24         y = px*y1 + py*x1;
25         printf("%10d%10d\n",y,(x-1)/2);
26         px = x;
27         py = y;
28     }
29     return 0;
30 }

View Code

poj 1320

标签：

原文地址：http://www.cnblogs.com/lmlyzxiao/p/4937781.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行