Codeforces Round #338 (Div. 2) E. Hexagons 找规律

题意：按题目给出的搜索方式给出第n个点的坐标。

思路：首先确定第n个点在哪一层？第k层有6k个点，所以前k层总共有3*k*（k+1）个点。二分找出n在哪一层之后。就可以把这一层分成6块，找出n在一块中，然后根据每一块相邻点的变化规律就可以找到第n个点的坐标。

http://codeforces.com/contest/615/problem/E

/*********************************************
    Problem : Codeforces
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++) //遍历
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --) //反向遍历
#define repS(it,p) for(auto it = p.begin() ; it != p.end() ; it ++) //遍历一个STL容器
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next) //遍历u所连接的点
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,m,k;
LL n;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

LL ef() {
    LL left = 1 , right = 2e9;
    LL mid;
    while(left < right) {
        mid = (left + right) >> 1;
        if((LL)3*mid*(mid+1) < n) {
            left = mid + 1;
        }
        else {
            right = mid ;
        }
    }
    return left;
}

void input() {

}

void solve() {
    if(n == 0) {puts("0 0"); return ;}
    LL pos = ef();
    LL now = n - (LL)3 * (pos-1) * pos;
    LL mark ;
    now ++;
    if(now == (LL) (pos * 6 + 1)) now = 1;
    rep(i,1,7) {
        if(now <= (LL)i * pos) {
            mark = i;
            break; 
        }   
    }
    LL x,y;
    now -= (mark-1) * pos;
    now -= 1;
    switch(mark) {
        case 1:x = 2 * pos ; y = 0 ; x -= now ; y += 2 * now; break;
        case 2:x = pos ; y = 2 * pos ; x -= 2 * now ;  break;
        case 3:x = -pos ; y = 2 * pos ; x -= now ; y -= 2 * now;break;
        case 4:x = -2 * pos ; y = 0 ; x += now ; y -= 2 * now;break;
        case 5:x = -pos ; y = -2 * pos ; x += 2 * now ; break;
        case 6:x = pos ; y = -2 * pos ; x += now ; y += 2 * now; break;
    }
    printf("%I64d %I64d\n",x,y);
}

int main(void) {
    //freopen("a.in","r",stdin);
    while(~scanf("%I64d",&n)) {
        input();
        solve();
    }
    return 0;
}