Street Numbers POJ

题意：就是从n到1再从1到n的各个数字之和为sum1, 然后从n到m，再从m到n的各个数字之和为sum2，求，（n,m）的前10组解。

思路：　　　直接建模，利用等差数列的求和公式计算一个公式（2n+1）^2 - m^2=1;   然后直接佩尔方程式即可！

#include<cstdio>
#include<cmath>
#define ll long long
int main()
{
    int x, y, x1, y1, px, py;
    x1 = 3;    y1 = 1;
    px = 3; py = 1;
    for (int i = 0; i < 10; ++i)
    {
        x = px*x1 + 8 * py*y1;    y = py*x1 + px*y1;
        printf("%10d%10d
", y, (x - 1) / 2);
        px = x; py = y;
    }
}

#include<cstdio>
#include<cmath>
#define ll long long
int main()
{
    int x, y, x1, y1, px, py;
    x1 = 3;    y1 = 1;
    px = 3; py = 1;
    for (int i = 0; i < 10; ++i)
    {
        x = px*x1 + 8 * py*y1;    y = py*x1 + px*y1;
        printf("%10d%10d
", y, (x - 1) / 2);
        px = x; py = y;
    }
}

#include<cstdio>
#include<cmath>
#define ll long long
int main()
{
    int x, y, x1, y1, px, py;
    x1 = 3;    y1 = 1;
    px = 3; py = 1;
    for (int i = 0; i < 10; ++i)
    {
        x = px*x1 + 8 * py*y1;    y = py*x1 + px*y1;
        printf("%10d%10d
", y, (x - 1) / 2);
        px = x; py = y;
    }
}