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

题意:

给定一个六边形坐标系统,从(0,0)出发,走环,求N≤1018步的坐标

分析:

显然,路径是一个环套一个环,下一个环插入了6个新的六边形,f(n)=f(n−1)+6,f(0)=1
首先先二分确定之前有多少环,然后求余数r
每个环的走法,都是6个方向,每个方向的步数是x(第x个环)
假设从(2x,0)出发,作为虚拟的第0步,我们可以通过变换规律求得r步的坐标
当然如果r=0,坐标就是(2×(x−1),0)

代码:

////  Created by TaoSama on 2016-01-09//  Copyright (c) 2015 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;LL n;int d[][2] = {{ -1, 2}, { -2, 0}, { -1, -2}, {1, -2}, {2, 0}, {1, 2}};int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%I64d", &n) == 1) {        LL l = 1, r = 1e9;        while(l <= r) {            LL m = l + r >> 1;            if(3 * m * (m + 1) <= n) l = m + 1;            else r = m - 1;        }        n -= 3 * l * (l - 1);        if(!n) {printf("%I64d 0\n", l - 1 << 1, 0); continue;}        LL x = l << 1, y = 0;        for(int i = 0; i < 6; ++i) {            LL k = min(n, l);            x += k * d[i][0];            y += k * d[i][1];            n -= k;        }        printf("%I64d %I64d\n", x, y);    }    return 0;}