POJ - 1320 - Street Numbers - （佩尔方程）

最新推荐文章于 2019-03-21 19:16:25 发布

原创最新推荐文章于 2019-03-21 19:16:25 发布 · 373 阅读

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#ACM #poj #数论 #佩尔方程

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本文介绍了一个有趣的数学问题，即寻找街道上特定编号的房子，使得从该房子出发向两个不同方向行走所经过的门牌号码之和相等。文章详细解释了如何利用佩尔方程来解决这一问题，并提供了完整的C++实现代码。

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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

题意是有一条街编号从1~n排列，一个人从门口出发往两边走，经过的门牌号的和相同，即若此人住在k号有题意1+2+3+...+k=k+(k+1)+(k+2)+...+n,让求k，n

用佩尔方程解，看到的不错的解释先贴上：（链接http://blog.youkuaiyun.com/u010885899/article/details/46765007），https://www.cnblogs.com/devymex/archive/2010/09/07/1818983.html

1+2+...+x = x+(x+1)+(x+2)...+y
求x，y。
方程就变为x*(x+1)/2 = (x+y)(y-x+1)/2 => y^2+y-2*x^2=0 => (2*y+1)^2-8*x^2=0
做这道题的收获就是通过这道题了解了佩尔方程，它是一个解x^2-d*y^2=1这类方程的方法。
佩尔方程的意思就是x^2-d*y^2=1的第一个解x0，y0已知的话，其余的值有一个递推公式了：
X(n)=X(n-1)*x0+d*Y(n-1)*y0
Y(n)=X(n-1)*y0+Y(n-1)*x0

以后记住解x^2-d*y^2=1的方程有一个简便算法~

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<queue>
#include<cstring>
#include<map>
using namespace std;
typedef long long ll;


int main()
{
    int i,x0=3,y0=1,x,y,prv_x=3,prv_y=1;
    for(i=1;i<=10;i++)
    {
        x=prv_x*x0+8*prv_y*y0;
        y=prv_x*y0+prv_y*x0;
        cout<<setw(10)<<y<<setw(10)<<(x-1)/2<<endl;
        prv_x=x;
        prv_y=y;
    }
    return 0;
}