Google笔试(2015年8月)

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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可以在这里下载
https://github.com/ncepuzhengyi/jobHuntingExam/tree/master/jobExam/Problem_20150815_google

Problem 1:
问题1是目前需要将阻止分成两块，由于组织内有些人之间有矛盾不能分到同一组内，问你是否存在这样的划分。
问题一是二分图判断问题，只需要判断无向图是否是二分图即可。最简单的方法是采用广度优先搜索+染色法即可。

#include <iostream>#include <string>#include <fstream>#include <map>#include <deque>using namespace std;#define size 5int 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;        else            cout << "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 again    int 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类似于插入排序的排序机制，对于接下来的元素，需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 again    int 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

下面这个是另外一次google的笔试题目，只收集到了这一道
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 logfrom math import ceildef Nlen(n):    return pow(2,n)-1def LineNum(k):    return ceil(log(k+1,2))r=Truedef func(x):    global r    lastNum=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=False    else:        r=True    return func(pow(2,lastNum)-x)s4=''for i in range(1,16):    r=True    s4+=func(i)print(s4)

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