codeforces 615E Hexagons (二分+找规律)

E. Hexagons

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ayrat is looking for the perfect code. He decided to start his search from an infinite field tiled by hexagons. For convenience the coordinate system is introduced, take a look at the picture to see how the coordinates of hexagon are defined:

Ayrat is searching through the field. He started at point (0, 0) and is moving along the spiral (see second picture). Sometimes he forgets where he is now. Help Ayrat determine his location after n moves.

Input

The only line of the input contains integer n (0 ≤ n ≤ 1018) — the number of Ayrat's moves.

Output

Print two integers x and y — current coordinates of Ayrat coordinates.

Sample test(s)

input

3

output

-2 0

input

7

output

3 2

题意：给你一个数n，是走的步数，问走了这么多步最后落点的坐标是多少；

思路：可以看出这些轨迹形成了一层一层的六边形，间距为2，先二分找出给的n最后落在第几层六边形上，再根据层数找到6个顶点跟层数的关系，最后判断最终位置的坐标，注意，在x的非负半轴最后要减2，因为一开始判断的时候这些点都判断多了一层；

AC代码：

#include <bits/stdc++.h>
using namespace std;
long long ansx,ansy,pos,n;
long long bisearch()
{
    long long l=0,r=1e9,mid;
    while(l<=r)
    {
        mid=((l+r)>>1);
        if(3*mid*(mid+1)>n)r=mid-1;
        else l=mid+1;
    }
    return l;
}
int check(long long x,long long y)
{
    if(x==0)ansx=2*pos-y,ansy=2*y;
    else if(x==1)ansx=pos-2*y,ansy=2*pos;
    else if(x==2)ansx=-pos-y,ansy=2*pos-2*y;
    else if(x==3)ansx=-2*pos+y,ansy=-2*y;
    else if(x==4)ansx=-pos+2*y,ansy=-2*pos;
    else if(x==5)ansx=pos+y,ansy=-2*pos+2*y;
    return 0;
}
int main()
{
    cin>>n;
    pos=bisearch();
    n-=3*pos*(pos-1);
    check(n/pos,n%pos);
    if(ansx>0&&ansy==0)ansx-=2;
    cout<<ansx<<" "<<ansy<<"\n";
    return 0;
}