Street Numbers
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 3078   Accepted: 1725

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
题目意思也就是,从家里向右走走到街道尾部经过的编号之和(不包括自己家)和从家里向左走走到街头经过的编号之和(不包括自己家)相等
设自己家的编号为m,街道共有n家,所以有

aaarticlea/gif;base64,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" alt="" />

化简整理得

aaarticlea/gif;base64,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" alt="" />

配凑得aaarticlea/gif;base64,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" alt="" />

aaarticlea/gif;base64,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" alt="" />

令aaarticlea/gif;base64,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" alt="" />,所以转化为佩尔型方程aaarticlea/gif;base64,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" alt="" />,且首项为aaarticlea/gif;base64,R0lGODlhJgASALMAAP///wAAAO7u7piYmGZmZoiIiMzMzDIyMkRERNzc3BAQECIiInZ2dlRUVKqqqrq6uiH5BAEAAAAALAAAAAAmABIAAASwEEgxpL04azBEJlsoTuBVGNcwFE01AgLyYE9xHRfjCoG7JbrADIOTqC4EhEXReAGEmYJDgkhseD4RFGNoAhQhhtK5xYC/Ggeh4SEPMQFAYhEyKFCv8mUhmIsaZyN6FnwCgRINeAAMAVaCbxdnhwkBU4hxeZBLVI4SDG2GNnJsG4MAXUZZAAUMLJanCqoCYgEHDIqrljFOEq9ORZ64Ir4jD6olIwVtLwLIFhQvxCIdFhEAOwA=" alt="" />

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
ll x,y,fx,fy,a,b;
int main()
{
x=fx=;y=fy=;
for(int i=;i<;i++)
{
a=fx*x+*fy*y;
b=fy*x+fx*y;
printf("%10lld%10lld\n",b,(a-)/);
fx=a;
fy=b;
}
return ;
}
												

POJ 1320 Street Numbers(佩尔方程)的更多相关文章

  1. POJ 1320 Street Numbers Pell方程

    http://poj.org/problem?id=1320 题意很简单,有序列 1,2,3...(a-1),a,(a+1)...b  要使以a为分界的 前缀和 和 后缀和 相等 求a,b 因为序列很 ...

  2. POJ 1320 Street Numbers 【佩尔方程】

    任意门:http://poj.org/problem?id=1320 Street Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Su ...

  3. POJ 1320 Street Numbers 解佩尔方程

    传送门 Street Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2529   Accepted: 140 ...

  4. POJ 1320:Street Numbers

    Street Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2753   Accepted: 1530 De ...

  5. POJ1320 Street Numbers【佩尔方程】

    主题链接: http://poj.org/problem?id=1320 题目大意: 求解两个不相等的正整数N.M(N<M),使得 1 + 2 + - + N = (N+1) + - + M.输 ...

  6. HDU 3292 【佩尔方程求解 && 矩阵快速幂】

    任意门:http://acm.hdu.edu.cn/showproblem.php?pid=3292 No more tricks, Mr Nanguo Time Limit: 3000/1000 M ...

  7. Poj 2247 Humble Numbers(求只能被2,3,5, 7 整除的数)

    一.题目大意 本题要求写出前5482个仅能被2,3,5, 7 整除的数. 二.题解 这道题从本质上和Poj 1338 Ugly Numbers(数学推导)是一样的原理,只需要在原来的基础上加上7的运算 ...

  8. 2010辽宁省赛G(佩尔方程)

    #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm& ...

  9. POJ 1320

    作弊了--!该题可以通过因式分解得到一个佩尔方程....要不是学着这章,估计想不到.. 得到x1,y1后,就直接代入递推式递推了 x[n]=x[n-1]*x[1]+d*y[n-1]*y[1] y[n] ...

随机推荐

  1. apiCloud中实现头部与内容分离与操作规范,App头部header固定,头部与内容分离

    官方案例 1.头部拆分成一个页面比如news-text <!doctype html> <html> <head> <meta charset="u ...

  2. 在vmware下为oracle RAC 创建共享存储的总结

    首先,介绍下用命令行vm-diskmanager形式创建磁盘文件的方法(其实,图形界面添加新磁盘就是调用此命令).       很多网上文章提及plainmaker.exe去创建共享磁盘,是以前的版本 ...

  3. ubuntu12.04下CKermit与开发板交互环境搭建

    CKermit蛮好的一个调试工具!就像在windows下的telnet,但是还是折腾了一下,现在看来,非常容易,其实我主要是在开发板为正常工作的情况下,以为是CKermit的问题,其实是我开发板开机设 ...

  4. ie浏览器检测不到cookie的问题

    之前做项目由于客户的要求设置缓存必须由后台来设置必须使用cookie(session是没问题的),后期设置时出现了登录页面与首页来还跳转的局面.原因就是首页没检测到登录后的缓存,而后台验证到确实已经是 ...

  5. Aspose.Words 操作指北

    前言:   这段时间因为业务需求,需要做一个word 导出,表单式的那种,因为之前也做过NPOI 操作 Excel 这类型的问题,所以总以为word 导出应该挺简单的,一直没有研究这块,汗颜,天真大意 ...

  6. Sql Server创建外键失败

    问题: 已成功保存“PPR_BasicInformation”表“PPR_PS”表- 无法创建关系“FK_PPR_PS_PPR_BasicInformation”. ALTER TABLE 语句与 F ...

  7. [转]Adobe CC 2018 下载链接 Creative Cloud 2018 - Creative Cloud 2018 – Adobe CC 2018 Download Links

    Creative Cloud 2018 – Adobe CC 2018 Download Links – ALL Languages Adobe CC 2018Direct Downloads Win ...

  8. <ItemTemp>里写判断语句

    <%@ Language="C#" %> <html> <head></head> <body> <%=DateT ...

  9. asp.net 连接字符串的多种写法

    一.使用OleDbConnection对象连接OLE DB数据源 1.连接Access 数据库 Access 2000: “provider=Microsoft.Jet.Oledb.3.5;Data ...

  10. maven/ssm框架搭建

    好久没有写java了,昨天学了下maven,不用手动的下载和添加jar包,实在是太方便. ------------------------------------------------------- ...