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@meaux, that code takes advantage of stream manipulators to format the output buffer: with setw(n) you define the width of the field to display, setfill(' ') lets you choose a padding character, and string(i,'#') builds a string consisting of i times the '#' character.
I had also gone for a one liner inside the loop (not really, as I had to reset a variable on every loop iteration) using std::transform and lambdas on a char array, but this is way more elegant and readable.
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
String rstr="";
for(int i=0;i<n;i++){
rstr+="#";
System.out.format("%"+n+"s%n",rstr);
}
}
The %n specifier does not print anything and the corresponding argument must be a pointer to signed integer where the number of characters written so far is stored.
int count;
printf("Jashin is a good coder %n,but not Best. ",&count);
/*
now compiler with count the number of character before %n and
will assign to count variable.*/
Scanner in = new Scanner(System.in);
int n = in.nextInt(),i=0;
while(++i <= n)
System.out.println(new String(new char[n-i]).replace("\0", " ")+new String(new char[i]).replace("\0", "#"));
Staircase
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int main(){ int n, space, hash; scanf("%d",&n);
}
How about
Nuh-uh, how about this. C++, ftw.
could you explain what the code is doing there?
this is magic to me.
Good one!! :)
@meaux, that code takes advantage of stream manipulators to format the output buffer: with setw(n) you define the width of the field to display, setfill(' ') lets you choose a padding character, and string(i,'#') builds a string consisting of i times the '#' character.
I had also gone for a one liner inside the loop (not really, as I had to reset a variable on every loop iteration) using std::transform and lambdas on a char array, but this is way more elegant and readable.
Java
Thanks for sharing this neat solution!
Can you please teacch me how "%"+n+"s%n" works
The %n specifier does not print anything and the corresponding argument must be a pointer to signed integer where the number of characters written so far is stored.
int count;
printf("Jashin is a good coder %n,but not Best. ",&count);
/* now compiler with count the number of character before %n and will assign to count variable.*/
printf("%d",count); 23
noo
i tried the same.. and output is also looking same but getting the wrong answer. can anyone explain why is it so?
how about this in java
Very nice
setfill(' ') is a little redundant isnt it?
Pretty much what I did, except I use
widthand took the defaults for fill and right.impressive
say about printf("%*s", n-i, "#");
The number (n-i) denotes the amount of spaces (%*s instead of %s) preceeing the string "#". So this:
Will print 10 spaces before printing a string "Abc"
That solution is great.
for (int i = 0; i < n; i++) { for (int j=0; j =k) System.out.print("#"); else System.out.print(" ");
}
No, it will print 9 spaces and a string "Abc". so total 10
extrodinary mind blowing fantastic
Greate solution
That doesn't work. will print the correct number of spaces, but only one '#'
I need to use scanf and printf more. I'm too much acustomed to cout and cin
cool
Can you do this by using only one loop?