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
題解：找規律題，把整個三角形轉化成一個三維的坐標軸 
例：對于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）； 對于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）; 兩個點的最短距離=x、y、z坐標差的絕對值
代碼：
#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 ); }}新聞熱點
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