尊敬的读者您好。从采访到领先的IT公司，我们为您带来一系列有趣的新问题和新任务，使您一次又一次地高兴！顺便说一句，上一期的问题答案已经发布！
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，我们从印度公司MakeMyTrip的采访中收集了任务。问题
1. 10枚硬币拼图您被蒙住了眼睛，桌上放着10个硬币。您可以触摸硬币，但凭感觉无法分辨出硬币的上升方向。有人告诉您，正面朝上有5个硬币，背面朝上有5个硬币，但哪一个不是。
您能否以相同的抬头数制造两堆硬币？您可以多次翻转硬币。
传递10 . , , . , 5 , («») 5 , («») , — .
, ? .
2. 报纸拼图由16张大纸制成的报纸对折。报纸共有64页。第一页包含页1、2、63、64。
如果我们拿起包含页号45的页，则该页还包含哪些其他页？
传递16 , . 64 . 1, 2, 63, 64.
*
, 45. ?
任务
1. 矩阵转置Write a program to find transpose of a square matrix mat[][] of size N*N. Transpose of a matrix is obtained by changing rows to columns and columns to rows.
Input:
The first line of input contains an integer T, denoting the number of testcases. Then T test cases follow. Each test case contains an integer N, denoting the size of the square matrix. Then in the next line are N*N space separated values of the matrix.
Output:
For each test case output will be the space separated values of the transpose of the matrix
Constraints:
1 <= T <= 15
1 <= N <= 20
-103 <= mat[i][j] <= 103
Example:
Input:
2
4
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
2
1 2 -9 -2
Output:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 -9 2 -2
Explanation:
测试案例1：旋转后的矩阵为：1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4。
传递, mat[][] N*N. .
:
T, . T . N, . N*N .
:
:
1 < = T <= 15
1 < = N <= 20
-103 < = mat[i] [j] < = 103
:
:
2
4
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4
2
1 2 -9 -2
:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 -9 2 -2
:
1: : 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4.
2. 阶乘尾随零对于整数n，请在n中找到尾随零的数目！。
输入：
第一行包含一个整数“ T ”，表示测试用例的总数。在每个测试用例中，它都包含一个整数“ N ”。
输出：
在每行中输出问题的答案。
约束：示例：
输入：输出：
1 <= T <= 100
1 <= N <= 1000
1
9
1
传递n n!.
:
'T', . 'N'.
:
.
:
1 < = T <= 100
1 < = N < = 1000
:
:
1
9
:
1
3. 骑士的脚步Given a square chessboard of N x N size, the position of Knight and position of a target is given. We need to find out minimum steps a Knight will take to reach the target position.
Input:
The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer n denoting the size of the square chessboard. The next line contains the X-Y coordinates of the knight. The next line contains the X-Y coordinates of the target.
Output:
Print the minimum steps the Knight will take to reach the target position.
Constraints:
1<=T<=100
1<=N<=20
1<=knight_pos,targer_pos<=N
Example:
Input:
2
6
4 5
1 1
20
5 7
15 20
Output:
3
9
N x N . , , .
:
T, . T . n, . X-Y . X-Y .
:
, , .
:
1<=T<=100
1<=N<=20
1<=knight_pos,targer_pos<=N
:
:
2
6
4 5
1 1
20
5 7
15 20
:
3
9
1: .
:
2 . .
:
— , — .
1:
2:
1.
1:
2:
.
245- 46. , P, 65-p .
,
65-45 = 20
65-46 = 19
, — 19, 20, 45, 46.
1O(1) :
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
public static void main (String[] args)
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try{
int testCases = Integer.parseInt(br.readLine());
while(testCases-->0){
int size = Integer.parseInt(br.readLine());
int[] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
int num = 0;
int i=0;
int total = size*size;
while(num<total && i<size){
if(num >= size*(size-1)) {
System.out.print(arr[num] + " ");
i++;
num = i;
}
else {
System.out.print(arr[num] + " ");
num+=size;
}
}
System.out.println();
}
}
catch (Exception e){
e.printStackTrace();
}
}
}
2import math
for _ in range(int(input())):
n=math.factorial(int(input()))
print(len(str(n))-len(str(n).rstrip("0")))
3#include<bits/stdc++.h>
using namespace std;
void bfs(bool visited[20][20],int n,int x,int y,int X,int Y,int& count){
queue<pair<int,int>>q;
visited[x-1][y-1]=true;
q.push(make_pair(x,y));
q.push(make_pair(0,0));
while(q.size()>1){
x=q.front().first;
y=q.front().second;
if(q.front().first==0 && q.front().second==0){
q.pop();
count++;
q.push(make_pair(0,0));
continue;
}
if(q.front().first==X && q.front().second==Y){
return;
}
if(x>2 && y>1 && visited[x-3][y-2]==false){
visited[x-3][y-2]=true;
q.push(make_pair(x-2,y-1));
}
if(x>2 && y<n && visited[x-3][y]==false){
visited[x-3][y]=true;
q.push(make_pair(x-2,y+1));
}
if(x>1 && y<n-1 && visited[x-2][y+1]==false){
visited[x-2][y+1]=true;
q.push(make_pair(x-1,y+2));
}
if(x<n && y<n-1 && visited[x][y+1]==false){
visited[x][y+1]=true;
q.push(make_pair(x+1,y+2));
}
if(x<n-1 && y<n && visited[x+1][y]==false){
visited[x+1][y]=true;
q.push(make_pair(x+2,y+1));
}
if(x<n-1 && y>1 && visited[x+1][y-2]==false){
visited[x+1][y-2]=true;
q.push(make_pair(x+2,y-1));
}
if(x<n && y>2 && visited[x][y-3]==false){
visited[x][y-3]=true;
q.push(make_pair(x+1,y-2));
}
if(x>1 && y>2 && visited[x-2][y-3]==false){
visited[x-2][y-3]=true;
q.push(make_pair(x-1,y-2));
}
q.pop();
}
}
int main()
{
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int x,y,X,Y;
cin>>x>>y>>X>>Y;
int count=0;
bool visited[20][20]={false};
bfs(visited,n,x,y,X,Y,count);
cout<<count<<'\n';
}
return 0;
}