🚮 🍑 🎚️ 第35期：IT培训-领先公司的当前问题和挑战 ⏭️ 👩🏼‍🔬 🏡

嗨，嗨！我们再次从领先的IT公司的访谈中为您准备了一系列有趣的问题和任务！

问题将每周出现-敬请期待！该专栏由招聘机构Spice IT支持。

本周，我们从美国公司PayPal的采访中收集了任务。顺便说一句，上一期的问题答案已经发布。

问题

1. 囚犯和警察难题

警察决定惩罚囚犯，并要求他发表声明。囚犯应该发表这样的声明，以便他还活着。
如果警察认为该陈述为真，则将囚犯处以绞刑，如果该陈述为假，则将其处决。

传递

. , .
, , , . ?

2. 火柴拼图

如何用6个相同的火柴棒制作4个等边三角形？

传递

4 6 ?

任务

1. 两个数组的交集

Given two arrays A and B respectively of size N and M. The task is to print the count of elements in the intersection (or common elements) of the two arrays.
For this question, intersection of two arrays can be defined as the set containing distinct common elements between the two arrays.

Input:
The first line of input contains an integer T denoting the number of test cases. The first line of each test case is N and M, N is the size of array A and M is size of array B. The second line of each test case contains N input A[i].
The third line of each test case contains M input B[i].

Output:
Print the count of intersecting elements.

Constraints:
1 ≤ T ≤ 100 1 ≤ N, M ≤ 105 1 ≤ A[i], B[i] ≤ 105

示例：
输入：输出：说明：测试案例1：89是两个数组交集中的唯一元素。测试案例2： 3 4 5和6是两个数组的交集中的元素。测试案例3：由于没有一个元素是通用的，因此输出为-1。测试案例4：10是唯一位于两个数组交集中的元素。
4 5 3 89 24 75 11 23 89 2 4 6 5 1 2 3 4 5 6 3 4 5 6 7 4 4 10 10 10 10 20 20 20 20 3 3 10 10 10 10 10 10

1 4 0 1

传递

A B N M. , ( ) .
, .

:
T, . — N M, N- A, M- B. N A[i].
M B[i].

:
.

:

1 ≤ T ≤ 100
1 ≤ N, M ≤ 105
1 ≤ A[i], B[i] ≤ 105

:
:

4
5 3
89 24 75 11 23
89 2 4
6 5
1 2 3 4 5 6
3 4 5 6 7
4 4
10 10 10 10
20 20 20 20
3 3
10 10 10
10 10 10

:
1: 89 — .
2: 3 4 5 6 — .
3: , 0.
4: 10 — , .

2. “ n”行中之字形字符串的串联

给定字符串和行数“ n”。当以行之字形方式写入输入字符串时，打印通过串联n行形成的字符串。

示例：输入：输入的第一行包含测试用例的数量。T个测试用例的描述如下：每个测试用例的第一行包含字符串，第二行具有'n'行数。输出： 在每行中将n行以Z字形连接后，打印字符串。约束：示例： 输入：输出：
Input: str = "ABCDEFGH" n = 2 Output: "ACEGBDFH" Explanation: Let us write input string in Zig-Zag fashion in 2 rows. A___C___E___G __B___D___F___H Now concatenate the two rows and ignore spaces in every row. We get "ACEGBDFH" Input: str = "SPICEITRECRUITMENT" n = 3 Output: SEEINPCIRCUTETITRM Explanation: Let us write input string in Zig-Zag fashion in 3 rows. S_______E_______E_______I_______N __P___C___I___R___C___U___T___E___T ____I_______T_______R_______M Now concatenate the two rows and ignore spaces in every row. We get "SEEINPCIRCUTETITRM"

1 ≤ T ≤ 70 1 ≤ N ≤ size of string

2 qrrc 3 rfkqyuqfjkxy 2

qrcr rkyqjxfqufky

传递

‘n'. , n , .

:

:
str = "ABCDEFGH"
n = 2
: "ACEGBDFH"
: 2 .
A___C___E___G
__B___D___F___H
. "ACEGBDFH"

: SEEINPCIRCUTETITRM
str = "SPICEITRECRUITMENT"
n = 3
:
: 3 .
S_______E_______E_______I_______N
__P___C___I___R___C___U___T___E___T
____I_______T_______R_______M
. "SEEINPCIRCUTETITRM"

:
. t :
, 'n' — .

:
n .

:

1 ≤ T ≤ 70
1 ≤ N ≤

:
:

2
qrrc
3
rfkqyuqfjkxy
2

qrcr
rkyqjxfqufky

3. 解决数独

Given an incomplete Sudoku configuration in terms of a 9 x 9 2-D square matrix (mat[][]). The task to print a solved Sudoku. For simplicity you may assume that there will be only one unique solution.

Sample Sudoku for you to get the logic for its solution:

Input:
The first line of input contains an integer T denoting the no of test cases. Then T test cases follow. Each test case contains 9*9 space separated values of the matrix mat[][] representing an incomplete Sudoku state where a 0 represents empty block.

Output:
For each test case, in a new line, print the space separated values of the solution of the the sudoku.

Constraints:
1 <= T <= 10 0 <= mat[] <= 9

Example:
Input:
1 3 0 6 5 0 8 4 0 0 5 2 0 0 0 0 0 0 0 0 8 7 0 0 0 0 3 1 0 0 3 0 1 0 0 8 0 9 0 0 8 6 3 0 0 5 0 5 0 0 9 0 6 0 0 1 3 0 0 0 0 2 5 0 0 0 0 0 0 0 0 7 4 0 0 5 2 0 6 3 0 0

Output:
3 1 6 5 7 8 4 9 2 5 2 9 1 3 4 7 6 8 4 8 7 6 2 9 5 3 1 2 6 3 4 1 5 9 8 7 9 7 4 8 6 3 1 2 5 8 5 1 7 9 2 6 4 3 1 3 8 9 4 7 2 5 6 6 9 2 3 5 1 8 7 4 7 4 5 2 8 6 3 1 9

Explanation:
Testcase 1: The solved sudoku is:
3 1 6 5 7 8 4 9 2 5 2 9 1 3 4 7 6 8 4 8 7 6 2 9 5 3 1 2 6 3 4 1 5 9 8 7 9 7 4 8 6 3 1 2 5 8 5 1 7 9 2 6 4 3 1 3 8 9 4 7 2 5 6 6 9 2 3 5 1 8 7 4 7 4 5 2 8 6 3 1 9

9 x 9 2D (mat [] []). . , .

, :

:
T, . T . 9 * 9 mat [] [], , 0 .

:
.

:

1 <= T <= 10
0 <= mat[] <= 9

:
:

1
3 0 6 5 0 8 4 0 0
5 2 0 0 0 0 0 0 0
0 8 7 0 0 0 0 3 1
0 0 3 0 1 0 0 8 0
9 0 0 8 6 3 0 0 5
0 5 0 0 9 0 6 0 0
1 3 0 0 0 0 2 5 0
0 0 0 0 0 0 0 7 4
0 0 5 2 0 6 3 0 0

3 1 6 5 7 8 4 9 2 5 2 9 1 3 4 7 6 8 4 8 7 6 2 9 5 3 1 2 6 3 4 1 5 9 8 7 9 7 4 8 6 3 1 2 5 8 5 1 7 9 2 6 4 3 1 3 8 9 4 7 2 5 6 6 9 2 3 5 1 8 7 4 7 4 5 2 8 6 3 1 9

:
1: :

3 1 6 5 7 8 4 9 2
5 2 9 1 3 4 7 6 8
4 8 7 6 2 9 5 3 1
2 6 3 4 1 5 9 8 7
9 7 4 8 6 3 1 2 5
8 5 1 7 9 2 6 4 3
1 3 8 9 4 7 2 5 6
6 9 2 3 5 1 8 7 4
7 4 5 2 8 6 3 1 9

— , — ’
, , , .
, , , .

t=0
try:
    t=int(input().strip())
except:
    pass
if __name__ == "__main__":
    while t:
        t-=1
        n, m = map(int, input().strip().split())
        a = list(map(int, input().strip().split()))
        b = list(map(int, input().strip().split()))
        a.sort()
        count=0;x=0
        for i in a:
            if i==x:
                pass
            else:
                if i in b:
                    count+=1
                    x=i
        print(count)

#include<bits/stdc++.h>
using namespace std;
int main()
 {
	//code
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t;
	cin>>t;
	while(t--)
	{
	    string str;
	    cin>>str;
	    int row;
	    cin>>row;
	    vector<char> v[row];
	    int flag=1,k=0;
	    while(1)
	    {
	        if(flag==1 && k<str.length())
	        {
	            for(int i=0;i<row && k<str.length();i++)
	             { 
	                 v[i].push_back(str[k]);
	                 k++;
	             }
	           flag=0;
	        }
	        if(flag==0 && k<str.length())
	        {
	            for(int i=row-2;i>0 && k<str.length();i--)
	            {
	                v[i].push_back(str[k]);
	                k++;
	            }
	            flag=1;
	        }
	        if(k==str.length())
	         break;
	    }
	    string res="";
	    for(int i=0;i<row;i++)
	    {
	         for(char ch:v[i])
	         {
	             res+=ch;
	         }
	         //cout<<endl;
	    }
	    cout<<res<<endl;
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int grid[9][9];
int row[9][10];
int col[9][10];
int box[3][3][10];
bool flag;
vector<pair<int,int> >V;
void call(int I)
{
    if(V.size()==I)
    {
        flag=false;
        for(int i=0;i<9;i++)
        {
            for(int j=0;j<9;j++)
            cout<<grid[i][j]<<" ";
        }
        cout<<endl;
        return;
    }
    int x=V[I].first,y=V[I].second;
    int boxx,boxy;
    if(x<3)
    boxx=0;
    else if(x<6)
    boxx=1;
    else
    boxx=2;
    
    if(y<3)
    boxy=0;
    else if(y<6)
    boxy=1;
    else
    boxy=2;
    
    
    for(int i=1;i<=9 && flag;i++)
    {
        if(row[x][i]==0 && col[y][i]==0 && box[boxx][boxy][i]==0)
        {
            row[x][i]=1; col[y][i]=1; box[boxx][boxy][i]=1;
            grid[x][y]=i;
            call(I+1);
            row[x][i]=0; col[y][i]=0; box[boxx][boxy][i]=0;
            grid[x][y]=0;
        }
    }
    return;
}
int main()
 {
	int t;
	cin>>t;
	while(t--)
	{
	    for(int i=0;i<9;i++)
	    {
	        for(int j=0;j<9;j++)
	        grid[i][j]=0;
	    }
	    for(int i=0;i<9;i++)
	    {
	        for(int j=1;j<=9;j++)
	        {
	            row[i][j]=0;
	            col[i][j]=0;
	        }
	    }
	    for(int i=0;i<3;i++)
	    {
	        for(int j=0;j<3;j++)
	            for(int k=1;k<=9;k++)
	            box[i][j][k]=0;
	    }
	    flag=true;
	    V.clear();
	    for(int i=0;i<9;i++)
	    {
	        for(int j=0;j<9;j++)
	        {
	            cin>>grid[i][j];
	            if(grid[i][j]==0)
	            {
	                V.push_back(make_pair(i,j));
	                continue;
	            }
	            row[i][grid[i][j]]=1;
	            col[j][grid[i][j]]=1;
	            if(i<3)
	            {
	               if(j<3)
	               {
	                    box[0][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[0][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[0][2][grid[i][j]]=1;
	               }
	               
	            }
	            else if(i<6)
	            {
	                if(j<3)
	               {
	                    box[1][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[1][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[1][2][grid[i][j]]=1;
	               }
	            }
	            else
	            {
	                if(j<3)
	               {
	                    box[2][0][grid[i][j]]=1;   
	               }
	               else if(j<6)
	               {
	                   box[2][1][grid[i][j]]=1;
	               }
	               else
	               {
	                   box[2][2][grid[i][j]]=1;
	               }
	            }
	        }
	    }
	    call(0);
	}
	return 0;
}

第35期：IT培训-领先公司的当前问题和挑战

问题

任务

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