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 因为序列很特殊所以我们用数学方法就可以解决 : 求和:  a*(a-1)/2 = (a+1+b)(b-a)/2 化简: 2a2  = b2 + b 两边乘4,构造完全平方项 (2b+1)2 - 8a2  = 1 令 x = 2*b+1; y = a; 我们就得到了一个形如Pell方程x2 - Dy2  = 1…

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…

传送门 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…

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…

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

题目链接 :  http://poj.org/problem?id=2427 PELL方程几个学习的网址: http://mathworld.wolfram.com/PellEquation.html     wolfram的讲解 http://hi.baidu.com/aekdycoin/item/a45f7c37850e5b9db80c03d1     AC神的博客 http://blog.csdn.net/acdreamers/article/details/8529686    acdr…

佩尔方程x*x-d*y*y=1,当d不为完全平方数时,有无数个解,并且知道一个解可以推其他解. 如果d为完全平方数时,可知佩尔方程无解. 假设(x0,y0)是最小正整数解. 则: xn=xn-1*x0+d*yn-1*y0 yn=xn-1*y0+yn-1*x0 证明只需代入. 如果忘记公式可以自己用(x0*x0-d*y0*y0)*(x1*x1-d*y1*y1)=1 推. 这样只要暴力求出最小特解,就可以用快速幂求出任意第K个解. Street Numbers Time Limit: 1000MS…

hdu2281 输入一个 $N$,求最大的 $n$($n \leq N$)和 $x$,使得 $x^2 = \frac{1^2+2^2+...+n^2}{n}$. 分析: 将右边式子的分子求和化简,有:$x^2 = \frac{(n+1)(2n+1)}{6}$. 变换成:$(4n+3)^2-48x^2 = 1$. 这就是佩尔方程的形式,且样例给出了最小整数解(7, 1). 求出long long范围内的所有解(也就9个) #include<bits/stdc++.h> using namespa…

一.Pell方程 形如x^2-dy^2=1的不定方程叫做Pell方程,其中d为正整数,则易得当d是完全平方数的时候这方程无正整数解,所以下面讨论d不是完全平方数的情况. 设Pell方程的最小正整数解为x0,y0则它的第n个正整数解满足xn+yn根号d=(x0+y0根号d)^n吧啦吧啦巴吧啦吧啦吧啦吧啦吧啦吧啦吧啦吧啦吧啦吧啦吧啦吧啦拉巴拉巴拉,然后随便撸撸就撸出递推式子了……OI里一般都要求第n个解,所以先暴力出最小解,然后快速幂搞,,,,,,,,,,, ————————————————————…

裸的pell方程. 然后加个快速幂. No more tricks, Mr Nanguo Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 320    Accepted Submission(s): 207 Problem Description Now Sailormoon girls want to tell you a ancien…