华电北风吹
天津大学认知计算与应用重点实验室
日期：2015/8/21

这三道题目的PDF能够在这里下载
https://github.com/ncepuzhengyi/jobHuntingExam/tree/master/jobExam/Problem_20150815_google

Problem 1:
问题1是眼下须要将阻止分成两块，因为组织内有些人之间有矛盾不能分到同一组内。问你是否存在这种划分。

问题一是二分图推断问题，仅仅须要推断无向图是否是二分图就可以。

最简单的方法是採用广度优先搜索+染色法就可以。

#include <iostream>
#include <string>
#include <fstream>
#include <map>
#include <deque>
using namespace std;#define size 5
int graph[size][size];
int visited[size];
int color[size];bool  GraphJudge(int nodeNum)
{memset(visited, 0, sizeof(visited));memset(color, 0, sizeof(color));for (int k = 0; k < nodeNum; k++){if (visited[k] == 0){visited[k] = 1;color[k] = 1;deque<int> q;q.push_back(k);while (q.empty()==false){int i = q.front();q.pop_front();for (int j = 0; j < nodeNum; j++){if (graph[i][j] > 0 && i!=j){if (visited[j] == 0){q.push_back(j);visited[j] = 1;color[j] = 1 - color[i];}else{if (color[j] == color[i])return false;}}}}}}return true;
}int main(int argc, char* argv[])
{ifstream in(".\\input.txt");cin.rdbuf(in.rdbuf());int T;cin >> T;cin.ignore();for (int caseNum = 0; caseNum<T; caseNum++){int M;cin >> M;memset(graph, 0, sizeof(&graph));map<string, int> nameIndex;map<string, int>::iterator it;int nodecount = 0;for (int i = 0; i < M; i++){int start, end;string p;cin >> p;it = nameIndex.find(p);if (it == nameIndex.end()){nameIndex[p] = nodecount;nodecount++;}start = nameIndex[p];cin >> p;it = nameIndex.find(p);if (it == nameIndex.end()){nameIndex[p] = nodecount;nodecount++;}end = nameIndex[p];graph[start][end] = 1;}if (GraphJudge(nodecount))cout << "Case #" << caseNum + 1 << ":" << "Yes" << endl;elsecout << "Case #" << caseNum + 1 << ":" << "No" << endl;}system("pause");return 0;
}

Problem 2:
问题二确切的说应该算作一个高中物理题，给出斜抛初速度和距离，计算斜抛初始角度。

#include<iostream>
#include <math.h>
#include <iomanip>
#include <ostream>
#include <fstream>
using namespace std;int main(int argc, char* argv[])
{//ifstream in("C:\\Users\\zhengyi\\Desktop\\ConsoleApplication1\\Debug\\input.txt");//streambuf *cinbuf = cin.rdbuf(); //save old buf//cin.rdbuf(in.rdbuf()); //redirect std::cin to in.txt!//ofstream out("C:\\Users\\zhengyi\\Desktop\\ConsoleApplication1\\Debug\\out.txt");//streambuf *coutbuf = std::cout.rdbuf(); //save old buf//cout.rdbuf(out.rdbuf()); //redirect std::cout to out.txt!//std::cin.rdbuf(cinbuf);   //reset to standard input again//std::cout.rdbuf(coutbuf); //reset to standard output againint T;cin >> T;int k = 0;while (k < T){int V, D;cin >> V >> D;double theta = asin(9.8 * D / (V * V)) * 90 / 3.14159265;cout << "Case #" << k + 1 <<":"<< fixed << setprecision(7) << theta << endl;k++;}return 0;
}

Problem 3:
第三题操作过程是一种相似于插入排序的排序机制，对于接下来的元素。须要往前插入的话就耗费1$。以此计算总共花费。

#include <iostream>
#include <string>
#include <vector>
#include <ostream>
#include <fstream>
using namespace std;int main(int argc, char* argv[])
{//ifstream in("C:\\Users\\zhengyi\\Desktop\\ConsoleApplication1\\Debug\\input.txt");//streambuf *cinbuf = cin.rdbuf(); //save old buf//cin.rdbuf(in.rdbuf()); //redirect std::cin to in.txt!//ofstream out("C:\\Users\\zhengyi\\Desktop\\ConsoleApplication1\\Debug\\out.txt");//streambuf *coutbuf = std::cout.rdbuf(); //save old buf//cout.rdbuf(out.rdbuf()); //redirect std::cout to out.txt!//std::cin.rdbuf(cinbuf);   //reset to standard input again//std::cout.rdbuf(coutbuf); //reset to standard output againint N;cin >> N;for (int i = 0; i < N; i++){int n;cin >> n;int count = 0;cin.sync();string str;vector<string> v1;while (count<n) {getline(cin, str);v1.push_back(str);count++;}//v1.erase(v1.begin());int cost = 0;for (int k = 1; k < n; k++){string key = v1[k];int t = k - 1;if (t >= 0 && v1[t].compare(key) > 0){v1[t + 1] = v1[t];t--;cost++;}v1[t + 1] = key;}cout << "Case #" << i + 1 << ":" << cost << endl;}return 0;
}

三道题目的代码
https://github.com/ncepuzhengyi/jobHuntingExam

以下这个是另外一次google的笔试题目。仅仅收集到了这一道
Problem 4
S0 = ”
S1 = ‘0’
S2 = ‘001’
S3 = ‘0010011’
S4 = ‘001001100011011’
……
SN = SN + ‘0’ + not（reverse（ SN ））
reverse是字符串反转
not是0，1转换
求S1010000 的第k个数
small data set: k<105
large data set: k<1018

from math import log
from math import ceildef Nlen(n):return pow(2,n)-1def LineNum(k):return ceil(log(k+1,2))r=True
def func(x):global rlastNum=LineNum(x)if x==pow(2,lastNum-1):if r:return '0'else:return '1'if x==1:if r:return '1'else:return '0'if r:r=Falseelse:r=Truereturn func(pow(2,lastNum)-x)s4=''
for i in range(1,16):r=Trues4+=func(i)
print(s4)