【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：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <map> 7 #include <vector> 8 #include <stack> 9 #include <string> 10 #include <ctime> 11 #include <queue> 12 #define mem0(a) memset(a, 0, sizeof(a)) 13 #define mem(a, b) memset(a, b, sizeof(a)) 14 #define lson l, m, rt << 1 15 #define rson m + 1, r, rt << 1 | 1 16 #define eps 0.0000001 17 #define lowbit(x) ((x) & -(x)) 18 #define memc(a, b) memcpy(a, b, sizeof(b)) 19 #define x_x(a) ((a) * (a)) 20 #define LL long long 21 #define DB double 22 #define pi 3.14159265359 23 #define MD 10000007 24 #define INF maxNum 25 #define max(a, b) ((a) > (b)? (a) : (b)) 26 using namespace std; 27 struct Point { 28 int x, y, z; 29 Point(int n) { 30 int xx = (int)sqrt(n + 0.5); 31 if(xx * xx < n) xx++; 32 y = xx; 33 z = x = y; 34 int p = (y - 1) * (y - 1); 35 x -= (n - p) / 2; 36 p = y * y + 1; 37 z -= (p - n) / 2; 38 } 39 }; 40 int main() 41 { 42 //freopen("input.txt", "r", stdin); 43 int a, b; 44 while(~scanf("%d%d", &a, &b)) { 45 Point x(a), y(b); 46 int ans = abs(x.x - y.x) + abs(x.y - y.y) + abs(x.z - y.z); 47 printf("%d\n", ans); 48 } 49 return 0; 50 }
