# 1 22 111 2222 11111 make this code from for loop

+1 vote

answered Nov 29, 2019 by (140 points)

Following code is the solution for

1 22 111 2222 11111

Code in "C Programming Language"

#include <stdio.h>

int main()
{
int i,j, m=1, n=2;

for(i=1; i<=5; i++)
{
for(j=1; j<=i; j++)
{
if(i%2 == 1)
printf("%d", m);
else
printf("%d", n);
}
printf(" "); /* \n is replaced with space */
}

return 0;
}

Code in "Python Programming Language"

for i in range(1, 6):
for j in range(1, i+1):
if(i%2 == 1):
print(1, end="")
else:
print(2, end="")
print("", end=" ")

I assume that you are asking for the following output

1

22

111

2222

11111

If yes, then following code is the solution

written in "C Programming Language" :

#include <stdio.h>

int main()
{
int i,j, m=1, n=2;

for(i=1; i<=5; i++)
{
for(j=1; j<=i; j++)
{
if(i%2 == 1)
printf("%d", m);
else
printf("%d", n);
}
printf("\n");
}

return 0;
}

Written in "Python Programming Language" :

for i in range(1, 6):
for j in range(1, i+1):
if(i%2 == 1):
print(1, end="")
else:
print(2, end="")
print("\r")

Hope this helps..

answered Nov 29, 2019 by Chaitanya
public static void main(String[] args) {

int even = 2;
int odd = 1;

for(int i=1;i<=20; i++){
if(i%2 == 0){
for(int j=1; j<=i;j++ ){
System.out.print(even);
}

}
else{
for(int j=1; j<=i;j++ ){
System.out.print(odd);
}
}
}
}
answered Nov 30, 2019 by Wesley Savage
for i in range(5):
if(i % 2 == 0):
num = "1"
else:
num = "2"
string = num
for b in range (i):
string += num
print(string)
answered Nov 30, 2019 by (140 points)
#include <stdio.h>

int main()

{

int a=5,i,j;

for(i=1;i<=a;i++)

{

if(i%2!=0)

for(j=1;j<=i;j++)

printf("1");

else

for(j=1;j<=i;j++)

printf("2");

printf(" ");

}

}
answered Nov 30, 2019 by Nuthan Kumar
```count = 1
for i in range(1,6):
if((i%2) == 0):
print('2'*count)
else:
print('1'*count)
count = count + 1```
answered Nov 30, 2019 by anonymous

#include<stdio.h>
int main(void)
{
int n,i,j;
printf("enter the total number of such sequences you want:");
scanf("%d",&n);

for(j=1;j<=n;j++)
{
for(i=1;i<=2*j-1;i++)
printf("1");
printf("\t");
for(i=1;i<=2*j;i++)
printf("2");
printf("\t");
}
return 0;
}

answered Nov 30, 2019 by anonymous
n=int(input())
a="1"
b="2"
for i in range(1,n+1):
if(i%2==0):
print(b*i)
else:
print(a*i)
answered Nov 30, 2019 by anonymous
#include<iostream>

using namespace std;

int

main ()

{

int i, n, j;

cout << "enter number of times that you want to print\n";

cin >> n;

for (i = 1; i <= n; i++)

{

cout << " ";

for (j = 1; j <= i; j++)

{

if (i % 2 == 1)

{

cout << "1";

}

else

{

cout << "2";

}

}

}

}
answered Nov 30, 2019 by anonymous
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin<<n;
for(int i=0;i<n;i++)
{
if(i%2==0)
{
for(int j=0;j<i+1;j++)
{
cout<<"1";
}
cout<<" ";
}
else
{
for(int j=0;j<i+1;j++)
{
cout<<"2";
}
cout<<" ";
}
}
}
answered Nov 30, 2019 by anonymous

#include <stdio.h>

int main()

{

int num,i,j;

scanf("%d",&num);

for(i=1;i<=num;i++)

{

if(i%2==1)

{

j=1;

for(int k=1;k<=i;k++)

printf("%d",j);

}

else

{

j=2;

for(l=1;l<=i;l++)

printf("%d",j);

}

printf(" ");

}

return 0;

}