POJ 1320 Street Numbers（佩尔方程）

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

化简整理得

配凑得

令,所以转化为佩尔型方程，且首项为

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
ll x,y,fx,fy,a,b;
int main()
{
    x=fx=3;y=fy=1;
    for(int i=0;i<10;i++)
    {
        a=fx*x+8*fy*y;
        b=fy*x+fx*y;
        printf("%10lld%10lld\n",b,(a-1)/2);
        fx=a;
        fy=b;
    }
    return 0;
}

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POJ 1320 Street Numbers（佩尔方程）

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