POJ 1320 Street Numbers(佩尔方程)
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 3078 | Accepted: 1725 |
Description
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
Output
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,R0lGODlhYgAUALMAAP///wAAAKqqqpiYmHZ2drq6ulRUVNzc3BAQEMzMzDIyMu7u7mZmZoiIiCIiIkRERCH5BAEAAAAALAAAAABiABQAAAT+EMhJgWiE1c2791f2jSS5NJJBlGx7pmsrk4UiDfasb/Wd70DKQSIKBoeAonEJcCCZF4yg5WRWFo9Cy6C1CrqAxpTEtU4OhEEAPGokAE+jYXE2kNxwM2VdEggWByhMBnYSAgMjfoCCegB8E4cEBYcGBwcBmAExS5cOBQWbCQIahpWZmo0SjwAFQwcIUwhvqRIHDgEKdAALiLIwtBurYwUIwBUEB8S+WpcTCGwfDITT1CofqxMEhS0JD97f4IwTc+MO2YUJAcYV2BIKiOsH2xJVTWMNP+uO0AsBSNBmDjyoMFCVrgebSEirRi1hsC7EAKiRkCBOozIUGQWgcwkgsEc0CRgIEEXgiz4MGODdSENgo74FBB7gIjBLn44GBW3q3PGMnsedQD0sIKUkqNERAwakOTohAgA7AA==" 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
- 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 ...
- 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 ...
- 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 ...
- POJ 1320:Street Numbers
Street Numbers Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 2753 Accepted: 1530 De ...
- 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 ...
- 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 ...
- 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 ...
- 2010辽宁省赛G(佩尔方程)
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm& ...
- 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
- Residual Networks <2015 ICCV, ImageNet 图像分类Top1>
本文介绍一下2015 ImageNet中分类任务的冠军--MSRA何凯明团队的Residual Networks.实际上.MSRA是今年Imagenet的大赢家.不单在分类任务,MSRA还用resid ...
- bzoj4568: [Scoi2016]幸运数字(LCA+线性基)
4568: [Scoi2016]幸运数字 题目:传送门 题解: 好题!!! 之前就看过,当时说是要用线性基...就没学 填坑填坑: %%%线性基 && 神犇 主要还是对于线性基的运用和 ...
- js捕获页面回车事件
1.javascript版 document.onkeyup = function (e) { if (window.event)//如果window.event对象存在,就以此事件对象为准 e = ...
- Java序列化注意事项
当父类继承Serializeble接口时,所有子类可以被序列化 子类实现了Serializeble接口,父类没有,父类中的属性不能序列化(不报错,数据会丢失),但是在子类中属性仍能正确序列化 如果序列 ...
- js小知识 双叹号(!!)
!!:一般用来将后面的表达式强制转换为布尔值(boolean):true或者false; avascript约定规则为: false.undefinded.null.0.”” 为 false tr ...
- 转js resplace方法使用
作者: hezhiwu5#163.com 时间:2007-3-22 大家好!!今晚在华软G43*宿舍没什么事做,把javascript中replace方法讲解一下,如果讲得不对或不合理是情理之中 ...
- javaweb实现教师和教室管理系统 java jsp sqlserver
1,程序设计思想 (1)设计三个类,分别是工具类(用来写连接数据库的方法和异常类的方法).信息类(用来写存储信息的方法).实现类(用来写各种操作数据库的方法) (2)定义两个jsp文件,一个用来写入数 ...
- 紫书 习题8-5 UVa 177 (找规律)
参考了https://blog.csdn.net/weizhuwyzc000/article/details/47038989 我一开始看了很久, 拿纸折了很久, 还是折不出题目那样..一脸懵逼 后来 ...
- 紫书 习题8-4 UVa 11491 (贪心)
题意:给你一个数, 要求删去一些数字, 使得剩下的数字最大. 这道题用贪心解决. 大家想一想, 两个数比较大小, 肯定先比较第一位的数,然后依次比较第二位,以此类推. 既然我们要保证最后的数字最大, ...
- 【Henu ACM Round#24 C】Quiz
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 肯定是这样 先放k-1个,然后空1个,然后再放k-1个.然后再空1个.. 以此类推. 然后如果(n/k)*(k-1)+n%k> ...