hdu 1030 Delta-wave

Delta-wave

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5781    Accepted Submission(s): 2203

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. 

 

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

 

#include<iostream>#include<algorithm>#include<cmath>#define MAX(x,y) ((x)>(y)?(x):(y))#define MIN(x,y) ((x)<(y)?(x):(y))using namespace std;int f(int x1,int x2,int y1,int y2){//调用前须保证y1为奇数 if(y1<=y2&&y2<=y1+2*(x2-x1)){//(x2,y2)在(x1,y1)的左右两条斜线的内部 return 2*(x2-x1)-!(y2&1);//若为奇数则为相隔的行数的二倍 ，否则行数的二倍-1 }else if(y1>y2){//(x2,y2)在(x1,y1)左斜线的左侧 return 2*(x2-x1)+y1-y2;}else {//(x2,y2)在(x1,y1)右斜线的右侧 return 2*(x2-x1)+y2-y1-2*(x2-x1);}}int N,M;int main(){int x1,x2,y1,y2,ans1,ans2,i,j,k,m,n;while(cin>>M>>N){if(M>N)swap(M,N);//使M在N上方 /*求出M,N的坐标*/ x1=(int)(ceil(sqrt((double) M))); y1=M-(x1-1)*(x1-1);x2=(int)(ceil(sqrt((double) N)));y2=N-(x2-1)*(x2-1);ans1=ans2=0x7fffffff;if(!(y1&1)){//若y1为偶数，则取两项的较小者 if(y1>1)ans1=f(x1,x2,y1-1,y2);if(y1<2*x2-1)ans2=f(x1,x2,y1+1,y2);cout<<MIN(ans1,ans2)+1<<endl;}else{//否则直接求值 cout<<f(x1,x2,y1,y2)<<endl;}}return 0;}