Finding the series / Getting the output of this type:
* * * * * * * * * *
* * * * * * * * *
* * * * * * *
* * * * *
* * *
* *
//This code takes an input from the user, and verifies whether its a even no. or an odd no.. If it is an odd number gives the output, else(If it is a even no.) it prints the message on the screen that please enter an odd no.and doesn't prints the output.
#include<math.h>
using namespace std;
int main()
{
int size,left,right,count,mid;
cout<<"Enter the size of the series(width) :";
cin>>size;
if(size%2==0)
{
cout<<"Its even number, series won't be accurate, please give any odd number....";
return 0;
}
round(mid);
//size=9;
count=1;
left=right=mid=((size+1)/2);
for(int j=0;j<=((size+1)/2);j++)///or///for(int j=0;j<(size/2)+1;j++)
{
for(int i=0;i<=size;i++)
{
//cout<<"left = "<<left<<" right = "<<right<<" mid = "<<mid<<endl;
if((i<left)||(left==right)||(i>=right))
cout<<" * ";
else
cout<<" ";
}
left--;con
right++;
cout<<endl;
}
}
/***
Explanation:
1 2 3 4 5 6 7 8 9 10
* * * * * * * * * *
* * * * * * * * *
* * * * * * *
* * * * *
* * *
* *
============================
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
* * * * * * * * * * * * * * * * * * * * *1
* * * * * * * * * * * * * * * * * * * *2
* * * * * * * * * * * * * * * * * *3
* * * * * * * * * * * * * * * *4
* * * * * * * * * * * * * *5
* * * * * * * * * * * *6
* * * * * * * * * *7
* * * * * * * *8
* * * * * *9
* * * *10
* *11
Conclusion:
So, for a even numbered size its not possible to obtain a correct seriesand for a odd number, the no. of lines of series will be the exactly half the sizeEx: for the size of 9, we'll get 5 as middle number, so we get series of lines
**/
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