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RGB7543 - Утга олголт 3

NxN хэмжээтэй квадрат массивыг жишээн дээрх загвараар дүүргэ.

Input

Квадрат массивын хэмжээ натурал тоо N өгөгдөнө. N<10.

Output

Квадрат массивын элементүүдийг мөр мөрөөр нь хэвлэнэ.

Гэхдээ элемент бүрийг 3 хоосон зайд хойш нь шахаж хэвлэнэ. Жишээг сайн ажиглана уу.

Жишээн дээрх 5-ын тоо гэхэд урдаа, хойноо тус бүр 2 хоосон зайтай байна. Мөн 15-ын тоо урдаа, хойноо тус бүр 1 хоосон зайтай байна.

Example

Input:
5

Output: 5 4 3 2 1
6 7 8 9 10
15 14 13 12 11
16 17 18 19 20
25 24 23 22 21

Нэмсэн:Bataa
Огноо:2013-02-07
Хугацааны хязгаарлалт:1s
Эх кодын хэмжээний хязгаарлалт:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Програмчлалын хэлүүд:ADA95 ASM32 BASH BF C NCSHARP CSHARP C++ 4.3.2 CPP C99 CLPS LISP sbcl LISP clisp D ERL FORTRAN HASK ICON ICK JAVA JS-RHINO JULIA LUA NEM NICE OCAML PAS-GPC PAS-FPC PERL PHP PIKE PRLG-swi PYTHON PYPY3 PYTHON3 RUBY SCALA SCM guile ST TCL WHITESPACE

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2019-11-16 05:51:02
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2019-11-09 05:16:07
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2019-11-09 05:15:53
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2019-03-20 04:42:19
#include<iostream>
using namespace std;
main(){
int n,i,j,s=1,a[101][101]={0};
cin>>n;
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
if(i%2==1)
{a[i][j]=i*n-j+1;}
else
{a[i][j]=(i-1)*n+j;}
}
}
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
h
2018-03-06 11:04:55
#include <iostream>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

long long j,l,i,n,a[10001][10001];

int main() {
cin>>n;
l=1;
for(i=1 ; i<=n ; i++){
for(j=1 ; j<=n ; j++){
a[i][j]=l;
l++;
}
}
for(i=1 ; i<=n ; i++){
for(j=n ; j>=1 ; j--){
if(a[i][j]/10>=1)cout<<" "<<a[i][j];
else cout<<" "<<a[i][j];
}
cout<<endl;
if(i!=n){i++;
for(l=1 ; l<=n ; l++){
if(a[i][l]/10>=1)cout<<" "<<a[i][l];
else cout<<" "<<a[i][l];
}
cout<<endl;
}
}

return 0;
}
2016-03-10 10:16:52 lhagvaa
#include<stdio.h>
#include<stdlib.h>
int s,d,f,t,g,i,n;
int a[100][100];
int main()
{
scanf("%d",&n);
f=0;
for(i=1; i<=n; i++)
if (i%2==0)
{
for(f=1; f<=n; f++)
{
t++;
t=a[i][f];
}
}
else
{
for(f=n; f>=1; f--)
{
t++;
t=a[i][f];
}
}
for(f=1; f<=n; f++)
{
for(g=1; g<=n; g++)
printf("%2d ",a[f][g]);
printf("\n");
}
system("pause");
return 0;
}

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