Street Numbers

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3624

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 Output

         6         8
        35        49

方法一：

题意：有1~m间屋子，并按1到m排列，有一人a住在第n间屋子。现要输出从1加到n-1和从n+1加到m的和相同时的n和m。

     思路：考察佩尔方程    由题意得

1+2+……+n-1=n*(n+1)/2
         (n+1)+(n+2)+……+m=(m-n)*(m+n+1)/2    即n*(n+1)=(m-n)*(m+n+1)
         化简得：2*n^2=m^2+m --> 8*n^2=(2*m+1)^2 -1 --> (2*m+1)^2 -8*n^2=1
          令x=2*m+1,y=n,d=8
            事先可以知道其中的最小解：x=3,y=1 接下来就是迭代了。
            佩尔方程原型：X^n - D*Y^n = 1
            其中佩尔方程的迭代公式：Xn = Xn-1 * X1 + d * Yn-1 * Y1
                                     Yn = Xn-1 * Y1 + Yn-1 * X1

代码：

#include<stdio.h>
int main()
{
    int x,y,count=10;
    int x1=3,y1=1,prex=3,prey=1,d=8;
    while(count--)
    {
        x=prex*x1+d*prey*y1;
        y=prex*y1+prey*x1;
        printf("%10d%10d
",y,(x-1)/2);
        prex=x;
        prey=y;
    }
    return 0;
}

方法二：

题目实际就是给你两个数m、n，使得m、n之间的数之和与1~m之间的数之和相等（不包括m）。

即：1+2+...+m-1 == (m+1)+...+n; 因此我们可以套用求和公式，两边移项化简得：2*pow(m,2) = n*(n+1)。

因此我们可以得到：m = sqrt(n*(n+1)/2)。枚举n的值，从而判断所得到的m值是否为整数，若为整数则说明两边的值实际相等。得到输出满足条件的最小的十组m、n，注意格式“%10d”，打表输出。

代码如下：

求满足条件的m，n：

代码：

 

#include<stdio.h>
#include<math.h>
int main()
{
    
    long long _mid;
    double mid;
    for(int i=6; i<100000000; i++)
    {
        mid = (double)i*(i+1);
        mid /= 2;
        mid = sqrt(mid);
        _mid = mid;// 取mid的整数部分
        if(fabs(mid-(double)_mid) < 1e-10)//若整数部分与本身相等则说明符合条件
            printf("%10I64d%10d
", _mid, i);
    }
    return 0;
}

 

方法3：打表法

代码：

#include <stdio.h>
int main()
{
    printf("%10d%10d
", 6, 8);
    printf("%10d%10d
", 35, 49);
    printf("%10d%10d
", 204, 288);
    printf("%10d%10d
", 1189, 1681);
    printf("%10d%10d
", 6930, 9800);
    printf("%10d%10d
", 40391, 57121);
    printf("%10d%10d
", 235416, 332928);
    printf("%10d%10d
", 1372105, 1940449);
    printf("%10d%10d
", 7997214, 11309768);
    printf("%10d%10d
", 46611179, 65918161);
    return 0;
}

 另一代码：

#include<stdio.h>
int main()
{
   int c=0,t=1,i,j,temp;
   int x[15]={3},y[15]={2};
   while(c<10)
   {
      x[t]=x[t-1]*x[0]+2*y[t-1]*y[0];
      y[t]=x[t-1]*y[0]+y[t-1]*x[0];
      t++;
      c++;
   }
   for(i=1;i<11;i++)
      printf("%10d%10d
",y[i]/2,(x[i]-1)/2);
   //system("pause");
   return 0;
}