Delta-wave
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7539 Accepted Submission(s):
2914
Problem Description
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.
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
Source
Recommend
很简单的题，弄懂题意，很容易写。
题意：求给出的两个数字所在表中的位置的距离。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 using namespace std; 6 int abs(int a) 7 { 8 return a>0?a:-a; 9 } 10 int main() 11 { 12 int n,m,i,j,cm,cn,rm,rn,lm,ln; 13 while(~scanf("%d%d",&m,&n)) 14 { 15 cm=(int)ceil(sqrt(m));//ceil为向上取整函数“math.h” 16 cn=(int)ceil(sqrt(n)); 17 rm=(m-(cm-1)*(cm-1)-1)/2+1; 18 rn=(n-(cn-1)*(cn-1)-1)/2+1; 19 lm=(cm*cm-m)/2+1; 20 ln=(cn*cn-n)/2+1; 21 int sum=(int)abs(cm-cn)+abs(rm-rn)+abs(lm-ln); 22 printf("%d ",sum); 23 } 24 return 0; 25 }