programming-challenges Bee Maja (111204) 题解

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本题可以通过找规律解决，事实上每个格子的值可以通过相邻的两个格子算出来的。我画了一个大大的图，其实也花了不少时间。

#include <iostream>#include <sstream>#include <fstream>#include <string>#include <vector>#include <list>#include <queue>#include <map>#include <set>#include <stack>#include <assert.h>#include <algorithm>#include <math.h>#include <ctime>#include <functional>#include <string.h>#include <stdio.h>#include <numeric>#include <float.h>using namespace std;const int MAXNUM = 100000; struct Node {int i1, i2; Node() : i1(0), i2(0) {}Node(int ai1, int ai2) : i1(ai1), i2(ai2) {}};Node addNode(const Node &n1, const Node &n2, int level) {Node newNode; if (n1.i1 == level && n2.i1 == level) {newNode.i1 = level + 1; }else if (n1.i1 == -level && n2.i1 == -level) {newNode.i1 = -level - 1; }else if (n1.i1 > 0 || n2.i1 > 0) {newNode.i1 = max(n1.i1, n2.i1);}else {newNode.i1 = min(n1.i1, n2.i1);}if (n1.i2 == level && n2.i2 == level) {newNode.i2 = level + 1;}else if (n1.i2 == -level && n2.i2 == -level) {newNode.i2 = -level - 1;}else if (n1.i2 > 0 || n2.i2 > 0) {newNode.i2 = max(n1.i2, n2.i2);}else {newNode.i2 = min(n1.i2, n2.i2);}return newNode;}vector<Node> vNodes; int main() {vNodes.push_back(Node(0, 0));vNodes.push_back(Node(0, 1));vNodes.push_back(Node(-1, 1));vNodes.push_back(Node(-1, 0));vNodes.push_back(Node(0, -1));vNodes.push_back(Node(1, -1));vNodes.push_back(Node(1, 0));int minorLoop = 0; int lastMinorLoop = 0; while (vNodes.size() <= MAXNUM) {lastMinorLoop++; minorLoop = lastMinorLoop * 6; int startIndex = vNodes.size() - minorLoop; for (int i = 1; i <= minorLoop; i++) {int delta = i - 1; if (i == 1) {vNodes.push_back(addNode(vNodes[startIndex], vNodes.back(), lastMinorLoop));}else {vNodes.push_back(addNode(vNodes[startIndex + delta], vNodes[startIndex + delta - 1], lastMinorLoop));}if ((i) % lastMinorLoop == 0) {vNodes.push_back(addNode(vNodes[startIndex + delta],  vNodes[startIndex + delta], lastMinorLoop)); }if (vNodes.size() > MAXNUM) break;}}int nodeId = 0; while (cin >> nodeId) {cout << vNodes[nodeId - 1].i1 << " " << vNodes[nodeId - 1].i2 << endl;}return 0; }

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