A triangle field is numbered with successive integers in the way shown on the picture below.
The traveller needs to go from the cell with number M to the cell with number N. The traveller is able to enter the cell through cell edges only, he can not travel from cell to cell through vertices. The number of edges the traveller passes makes the length of the traveller’s route.
Write the PRogram to determine the length of the shortest route connecting cells with numbers N and M.
Input Input contains two integer numbers M and N in the range from 1 to 1000000000 separated with space(s).
Output Output should contain the length of the shortest route.
Sample Input 6 12
Sample Output 3
题解：找规律题，把整个三角形转化成一个三维的坐标轴 
例：对于6，在第三行，z=3， x=(9-6)/2+1=2, y=(6-4-1)/2+1=1,所以（2，1，3）； 对于12，在第4行，z=4, x=(16-12）/2+1=3，y=(12-9-1)/2+1=2,所以（3，2，4）; 两个点的最短距离=x、y、z坐标差的绝对值
代码：
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 1005using namespace std;int x[1000005];int main(){ int n, m; while (cin >> n >> m) { int Za = (int)ceil(sqrt(n * 1.0)); int Zb = (int)ceil(sqrt(m * 1.0)); int Xa = (Za * Za - n) / 2 + 1; int Xb = (Zb * Zb - m) / 2 + 1; int Ya = (n - (Za - 1) * (Za - 1) - 1) / 2 + 1; int Yb = (m - (Zb - 1) * (Zb - 1) - 1) / 2 + 1; int ans = (int)( fabs((Za - Zb) * 1.0) + fabs((Xa - Xb) * 1.0) + fabs((Ya - Yb) * 1.0) ); printf("%d/n", ans ); }}新闻热点
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