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RGB7401 - Давхар давталт 1

Өгөгдсөн тоо болон тэрхүү тоо хүртэлх тоонуудыг жишээн дээрх хэлбэрээр хэвлэ.

Input

Бүхэл тоо өгөгдөнө. 10-аас бага.

Output

Тоонуудыг зайгаар тусгаарлан жишээн дээрх хэлбэрээр мөрүүдэд хэвлэнэ.

Example

Input:
5

Output:
5 4 3 2 1
5 4 3 2 1
5 4 3 2 1
5 4 3 2 1
5 4 3 2 1

Нэмсэн:Bataa
Огноо:2013-01-14
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

hide comments
2025-02-12 03:43:45
#include <iostream>
#include <sstream>
#include <string>
#include <cstdlib>
#include <cmath>

using namespace std;

// Headers
string toString (double);
int toInt (string);
double toDouble (string);

int main() {
int i, j, n;

cin >> n;
for (i = 1; i <= n; i++) {
for (j = n; j >= 1; j--) {
cout << j;
cout << " ";
}
cout << "" << endl;
}
return 0;
}

// The following implements type conversion functions.
string toString (double value) { //int also
stringstream temp;
temp << value;
return temp.str();
}

int toInt (string text) {
return atoi(text.c_str());
}

double toDouble (string text) {
return atof(text.c_str());
}
2025-02-02 17:18:40
#include<iostream>
using namespace std;
int main (){
int a;
cin>>a;
for (int i=1 ;i<=a; i++){

for (int j=a; j>=1; j--)
cout<<j<<" ";
cout<<endl;
}
return 0;
}//zzo amjilt
2025-01-29 03:25:54
Bi dudud hairtai
2025-01-29 03:25:19
Dudu single azgu al
2025-01-29 03:25:06
Bi dudud hairtai
2025-01-29 03:24:27
dudu tng psdshuuu hrjivl hrj urdg shu <3
2025-01-29 03:24:17
Dudu temneg huja monkey

Last edit: 2025-01-29 03:24:49
2024-01-21 10:09:21
#include <stdio.h>
int n,i,j;
int main()
{
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=n;j>=1;j--)
printf(" %d",j);
printf("\n");
}
return 0;
}
2023-12-01 11:06:56
gomo amar
2023-03-15 02:15:18
#include<stdio.h>{COuS}
int main()
{
int n, i, j;
scanf("%d",&n);
for(i=n; i>=1; i--)
{
for(j=n; j>=1; j--)
printf(" %d",j);
printf("\n");
}
}
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