1 121 12321 1234321 program

Answer 1

1^2 = 1
11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321













#include
#include
#include
void main()
{
clrscr(); int n;//soullessgod
cout<<"\n enter number of lines ";//whatsoever
cin>>n;
int a=1,b,s=1;
for(;a<=n;s=s*10+1)
{
for(b=n-a;b>=1;b--)
{
cout<<" ";
}
cout<cout<<"\n";
a++;
}

getch();
}

Answer 2

for(int r=1; r<=5; r++)

{

for(int sp=5-r; sp>0; sp--)

System.out.print(" ");

for(int c=r; c>=1; c--)

System.out.print(c);

for(int c=2; c<=r; c++)

System.out.print(c);

System.out.print("\n");

}

for(int r=1; r<=5; r++)

{

for(int sp=r; sp>=1; sp--)

System.out.print(" ");

for(int c=5-r; c>=1; c--)

System.out.print(c);

for(int c=2; c<=5-r; c++)

System.out.print(c);

System.out.print("\n");

}

Answer 3

the series continues with the length of odd series ie.1,3,5,7...so on.the middle integer is extra sdded in every number.it's general structure is

(inital numbers)(extra number)(reverse of initial numbers)

extra number is always multplied by 2 with last extra number added

char s[50];

s[0]=1;

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

{

traverse the array to next middle position

add the extra element in between the elements

concatenate the reverse part of initial elements

}

this can b done as

for(i=1,j=0;i<n;i=strlen(s),j++)

{

for(k=i;k>j;k--)

{

s[k+1]=s[k];

}

s[j+1]=s[j-1]*2;

s[i+3]='\0';

print s;

}

Answer 4

#include <stdio.h>

int main (void)

{

puts ("12345 51234");

return 0;

}