【hdu1030】“坐标表示法”

http://acm.hdu.edu.cn/showproblem.php?pid=1030

算法：以顶点为原点，建立坐标系，一个数可以唯一对应一个三元组（x, y, z），从任意一个点出发走一步，刚好有三种情况，分别对应x, y, z变化1，而其它两个坐标保持不变。因此，求出两个点的坐标分别为(x1, y1, z1), (x2, y2, z2);则它们之间的距离为|x1 - x2| + |y1 - y2| + |z1 - z2|。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <vector>
#include <stack>
#include <string>
#include <ctime>
#include <queue>
#define mem0(a) memset(a, 0, sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define eps 0.0000001
#define lowbit(x) ((x) & -(x))
#define memc(a, b) memcpy(a, b, sizeof(b))
#define x_x(a) ((a) * (a))
#define LL long long
#define DB double
#define pi 3.14159265359
#define MD 10000007
#define INF maxNum
#define max(a, b) ((a) > (b)? (a) : (b))
using namespace std;
struct Point {
        int x, y, z;
        Point(int n) {
                int xx = (int)sqrt(n + 0.5);
                if(xx * xx  < n) xx++;
                y = xx;
                z = x = y;
                int p = (y - 1) * (y - 1);
                x -= (n - p) / 2;
                p = y * y + 1;
                z -= (p - n) / 2;
        }
};
int main()
{
        //freopen("input.txt", "r", stdin);
        int a, b;
        while(~scanf("%d%d", &a, &b)) {
                Point x(a), y(b);
                int ans = abs(x.x - y.x) + abs(x.y - y.y) + abs(x.z - y.z);
                printf("%d
", ans);
        }
        return 0;
}