Rhombus Pattern - N Slashes Side
Given an odd value of N, the program must print a rhombus in diamond shape whose side contains N slashes as shown below in the examples.
Input Format:
The first line contains N.
Output Format:
The rhombus in diamond shape with each side containing N slashes. Asterisk is used as a filler for other values.
Boundary Conditions:
1 <= N <= 101 and N is odd.
Example Input/Output 1:
Input:
5
Output:
****/\****
***/**\***
**/****\**
*/******\*
/********\
\********/
*\******/*
**\****/**
***\**/***
****\/****
Example Input/Output 2:
Input:
11
Output:
**********/\**********
*********/**\*********
********/****\********
*******/******\*******
******/********\******
*****/**********\*****
****/************\****
***/**************\***
**/****************\**
*/******************\*
/********************\
\********************/
*\******************/*
**\****************/**
***\**************/***
****\************/****
*****\**********/*****
******\********/******
*******\******/*******
********\****/********
*********\**/*********
**********\/**********
Input Format:
The first line contains N.
Output Format:
The rhombus in diamond shape with each side containing N slashes. Asterisk is used as a filler for other values.
Boundary Conditions:
1 <= N <= 101 and N is odd.
Example Input/Output 1:
Input:
5
Output:
****/\****
***/**\***
**/****\**
*/******\*
/********\
\********/
*\******/*
**\****/**
***\**/***
****\/****
Example Input/Output 2:
Input:
11
Output:
**********/\**********
*********/**\*********
********/****\********
*******/******\*******
******/********\******
*****/**********\*****
****/************\****
***/**************\***
**/****************\**
*/******************\*
/********************\
\********************/
*\******************/*
**\****************/**
***\**************/***
****\************/****
*****\**********/*****
******\********/******
*******\******/*******
********\****/********
*********\**/*********
**********\/**********
Code:
#include <iostream>
using namespace std;
int main(int argc, char** argv)
{
int n,i,j;
cin>>n;
int x,y,flag=1,flag1=1;
x=n-1;
y=n;
for(i=0;i<n;i++)
{
for(j=0;j<n*2;j++)
{
if(x==j&&flag1==1)
{
cout<<"/";
x--;
flag1=0;
}
else
if(y==j&&flag==1)
{
cout<<"\\";
y++;
flag=0;
}
else
cout<<"*";
}
cout<<"\n";
flag=1,flag1=1;
}
x=0;y=n*2-1;
flag=1;flag1=1;
for(i=0;i<n;i++)
{
for(j=0;j<n*2;j++)
{
if(x==j&&flag==1)
{
cout<<"\\";
x++;
flag=0;
}
else if(y==j&&flag1==1)
{
cout<<"/";
y--;
flag1=0;
}
else
cout<<"*";
}
cout<<"\n";
flag=1;flag1=1;
}
}
please correct your code;
ReplyDeletein second 'i' loop
for(i=0;i=2*n;i++)
Yolo...I made this by dividing the whole into two halves..
Deleteeg: when the n=5
First loop iterates 0 to n
Second loop iterates n to n*2