Mani S repliedinclude<stdio.h>
#include<conio.h>
void main()
{
int i,j;
clrscr();
for(i=0;i<=3;i++)
{
for(j=i+1;j>=1;j--)
{
printf("%d",j);
}
printf("n");
}
for(i=3;i>=1;i--)
{
for(j=i;j>=1;j--)
{
printf("%d",j);
}
printf("n");
}
}

Hope this helps. If it does, please select my answer as the best one!! Happy programming :)

Chie replied// n = maximum number to be printed + 1
int i;
int j;
......

for(i = 0; i < n; i++) {
____for(j = i + 1; j >= 1; j--)
________printf("%d", j);
____printf("n");
}
for(i = n - 2; i >= 0; i++) {
____for(j = i + 1; j >= 1; j--)
________printf("%d", j);
____printf("n");
}

Josh A repliedThis is for a console application and is in C++.



#include <iostream>
#include <stdlib.h>

int main(int argc, char *argv[])

cout << "1" << endl;
cout << "21" << endl;
cout << "321" << endl;
cout << "4321" << endl;
cout << "321" << endl;
cout << "21" << endl;
cout << "1" << endl;

system("PAUSE")
return 0;