Pell方程(求形如x*x-d*y*y=1的通解。)

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		Memory Limit: 10000K
Total Submissions: 2813		Accepted: 1568

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

这题可以得到佩尔方程s*s-8*t*t=1 ,s=2n+1,t=x (n表示总长，x表示取的n中某个位置)

s0=3,t0=1 然后就很好弄了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
using namespace std;

int main(int argc, const char * argv[]) {
    long long s0=;
    long long t0=;
    long long s1=;
    long long t1=;
    for(int i=;i<=;i++)
    {
        long long s,t;
        s=s1*s0+*t1*t0;
        t=t1*s0+t0*s1;
        s1=s;
        t1=t;
        printf("%10lld%10lld\n",t,(s-)/);
        //cout<<t<<" "<<(s-1)/2<<endl;
    }
    return ;
}

这题用暴力然后打表也是可以0MS过的。