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Number Pattern - 32

1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1


#include<stdio.h>
int main()
{
  int i,j;
  for(i=1;i<=4;i++)
  {
    for(j=1;j<=i;j++)
    {
      if(j<i) 
        printf("%d*",i);
      else 
        printf("%d",i);
    }
    printf(" \n");
  }
  for(i=4;i>=1;i--)
  {
    for(j=1;j<=i;j++)
    {
      if(j<i) 
        printf("%d*",i);
      else 
        printf("%d",i);
    }
    printf(" \n");
  }  
  return 0;
}


Related Links:
- More Number Pattern Programs
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6 comments:

  1. following is the small modification of above program
    for(i=1;i<=4;i++)
    {
    for(j=1;j<=i;j++)
    {
    if(j!=1)
    printf("%d*",i);
    else
    printf("%d",i);
    }
    printf(" \n");
    }

    ReplyDelete
    Replies
    1. //program to print the following pattern:

      // 1
      // 2 * 2
      // 3 * 3 * 3
      // 4 * 4 * 4 * 4
      // 4 * 4 * 4 * 4
      // 3 * 3 * 3
      // 2 * 2
      // 1


      #include
      int main()
      {
      int k=1, m=1;
      for(int i=1; i<=8; i++)
      {
      int x;
      if(i>=1&&i<=4)
      x=i;
      else //(i>=5&&i<=8)
      x= i-m;
      for(int j=1; j<=7; j++)
      {
      if(j>=1&&j<=k)
      {
      if(j%2==0)
      printf("* ");
      else
      printf("%d ", x);
      }
      }
      printf("\n");
      if(i<=3)
      k +=2;
      else if(i>=5)
      {
      k -=2;
      m +=2;
      }
      }
      return 0;
      }

      Delete
  2. #include

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

    printf("Enter the no. of rows : ");
    scanf("%d",&r);

    for(i=1;i<=r;i++)
    {
    for(j=1;j<=i;j++)
    {
    printf("%d",i);

    if(j!=i)
    {
    printf("*");
    }
    }

    printf("\n");
    }

    for(i=r;i>=1;i--)
    {
    for(j=1;j<=i;j++)
    {
    printf("%d",i);

    if(j!=i)
    {
    printf("*");
    }
    }

    printf("\n");
    }

    return 0;
    }

    ReplyDelete
  3. #include <stdio.h>
    void main()
    {
    int i,j;
    for(i=1;i<=4;i++)
    {
    for(j=1;j<=i;j++)
    { if(j==i)
    printf("%d",i);
    else
    printf("%d*",i);
    }


    printf("\n");

    }
    for(i=4;i>=1;i--)
    {
    for(j=i;j>=1;j--)
    { if(j==i)
    printf("%d",i);
    else
    printf("*%d",i);
    }


    printf("\n");

    }
    }

    ReplyDelete