Maycon_Ferreira As a C programmer
public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int N = in.nextInt(); for(int i = 1; i <= 10; i++){ System.out.printf("%d x %d = %d%n", N, i, N*i); } } }
LuQzz This is clean! Awesome!
VIPERCODER1 may i ask that where is the declaration of "i"?
Jordeatsu within the for loop (int i = 1; i<=10; i++)
VIPERCODER1 oh okay....i got it,thanks
TARUNMATTA1994 int i is the declaration of i
prem_vishwakarm1 its up to you define it in main function or inside the for loop its up to you. just dont forgrt to initialize it to 1
maniidamakanti_1 Why we use printf instead of println?
brendanford24 What have you imported for this code?
public class Solution { public static void main(String[] args) { Scanner input = new Scanner(System.in); for (int i = 1; i <= 10 && input.hasNextInt(); i++) { int n = input.nextInt(); System.out.printf("%d x %d = %d\n", n, i, (n*i)); } } }
Mine wont work^^?
brendanford24 my bad, i see where i kooked it
import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner input = new Scanner(System.in); int n = input.nextInt(); for (int i = 1; i <= 10; i++) { System.out.printf("%d x %d = %d\n", n, i, n*i); } } }
Now working ^^
chigurupatichai1 is it working
rmyoung04 Good call on using printf. However, you need to close scanner.
himaanshurocks you can simply write:
System.out.println(input+" x "+i+" = "+(2*i));
celik_zeki i did this but after i saw formatted print i am not contented with my code :(
nilupulee_hansa1 Here "2" should not be a constant. right? It is the input value. Isn't it?
zee13_khan Then it will be the output for just 2, here we are taking input number to generalize the code
asaxenarec123 why not ?? here 2 will be constant because here we have to only table of 2, you can see output of given instruction...
aakashkumarjee at the end you should write System.out.println(input+" x "+i+" = "+(input*i));
amittomar87 Could someone plz elaborate the use of "%d x %d = %d\n"? How did it manage to come as n * i?
kumar_santosh959 System.out.printf("%d x %d = %d%n", n, i, n*i); try this it work
Venkat30119 input.hasNextInt() purpose of using it??
himaanshurocks not sure . but it may be useful to check whether there exists a new int or not.
aakashkumarjee this method tells about the token present in te buffer. i.e it will return true if token is present and false if it is not. for example, take a look at this
public class ScannerDemo { public static void main(String[] args) {
String s = "Hello World! 3 + 3.0 = 6"; // create a new scanner with the specified String Object Scanner scanner = new Scanner(s); // check if the scanner has a token System.out.println("" + scanner.hasNext()); // print the rest of the string System.out.println("" + scanner.nextLine()); // check if the scanner has a token after printing the line System.out.println("" + scanner.hasNext()); // close the scanner scanner.close();} }
output of this program will be
true
Hello World! 3 + 3.0 = 6
false
ghazi261 what does it mean by token in this case? Can you please explain or refer to an explaination? Thanks
aakashkumarjee basically token is the smallest building block of any program. the scanner object may or may not contain any token, hence first we confirm it.
ganeshethiraj185 util package is enough
ravi_bhatt_75491 it should be \n instead of %n
Maycon_Ferreira no it shouldn't
"A new line character appropriate to the platform running the application. You should always use %n, rather than \n."
https://docs.oracle.com/javase/tutorial/java/data/numberformat.html
deekshithsagar73 yes it should
vjitesh Will you explain me: System.out.printf("%d x %d = %d%n", N, i, N*i);
sara99 [deleted]deekshithsagar73 NOO
karthikshenoy93 Printf is used to format a output as required; %d is for int, similarly we have %s for String and %f for float and so on. Also we have %n for the new line which serves the purpose of println for printf. Here the first %d is matched with the first value after the ',', if the format dosent match exception is thrown. %d --> N %d --> i %d --> N*i
tasfiahm Thanks a lot. Today I have lerned something new and simple but very inportant in fromating lines.
rgdeekshith can you please explain sir why did you use "printf" function instead of "println" function and kindly could you tell me its purpose?
Maycon_Ferreira [deleted]deekshithsagar73 NOO
kacha_nimbu public static void main(String[] args) { int N = scanner.nextInt(); int iLoop; //scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for(iLoop=1;iLoop<11;iLoop++) { int ans= N*iLoop; System.out.println(N+" x "+iLoop+" = "+ans); } scanner.close(); you can use it like this
Pricks_123 Exactly Correct.Thanks for uploading solution
deekshithsagar73 thanks for uploading comment
abhijitaj123 i m facing this issue, can u pls help me?
error: no suitable method found for println(String,int,int,int)
Pricks_123 can u share ur entire code, so that we could analize ur issue
deekshithsagar73 noo
kumchandan07 Use printf(String,int,int,int) instead. There is no overloaded method of println having arguments as String,int,int,int. Incorrect: println(String,int,int,int)
sownderya29 can u explain this!!without getting a value how this get printed and ya also you are using printf here
imvikash007k I Need a Help Regarding N limit N>=2&&N<=20 No one Gave The range of N so for this challenge if n=30 the following answer will not work.
3333ayannkhan can you explain this : %d x %d
VIPERCODER1 Yes ofcourse my friend, the places where you see %d, you will have to presume that the value of the element that is needed to be assigned will replace it. we will just have to keep in mind its syntax that is sopf( %d * %d ", a, b); to make you more clear i will also say that in the above line the first %d will replace it with the first element after thye double quotes that is a. thank you :)
manojred221 will this pass the test case with N=45
singhsaurabh_ki1 how i print x ??? shift +8=*.. but its wrong????why?
umarbasha007 private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for(int i=1;i<=10;i++) { // Can use \n instead of %n, both works fine System.out.printf("%d x %d = %d\n",N,i,N*i); } scanner.close(); }Saitama_sensei what does scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); do???
kacha_nimbu Couldn't find much about it. My code is working fine with and without
ishraqisc Why aren't you setting up the constraint for N, i.e. 2<=N<=20?
shivayush143 in the for loop instead of using only i y do u use integer type i. u cn show it in above.
shivayush143 does it take the value of N throgh user
ankushchandole what is the diffrence between these two lines
System.out.println("%d x %d = %d ",N,i,N*i); &
System.out.printf("%d x %d = %d%n", N, i, N*i);whawkins90 The printf method expects a formatted String as its first argument and will substitute the values of the subsequent parameters into the String. The println method expects a non-formatted String; it will print exactly what you pass into it and will not perform any substitution.
momsteachcode Why can't we use println here instead of printf, why is it showing an error
krishnadulaldal1 [deleted]tigerO [deleted]
spacetaco94 for(int i=0; i<10; i++){
System.out.printf("%d x %d = %d\n", N, (i+1), (N*(i+1)));}
Emmanuelmireku15 This is what I got but after reading the discussions, I a lot of people using the printf method. But this code here will do the same job. This code is in Java 8.
import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner sc = new Scanner(System.in); int input = sc.nextInt(); sc.close(); for(int i = 1; i <= 10; i++){ int answer = input*i; System.out.println(input + " x " + i + " = " + answer); } } }
lochek_alon this is exactly how i solved it, but need to keep using
printf, i think, since everyone uses it here.todd37 YES, the directions indicate you must use printf (format printing).
contactredch can't understand why mine gives the error. public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
for(int i=1; i<=10;i++){ int z = N*i; } System.out.println(N + " x " + i + " = " + z); scanner.close(); }
whawkins90 Looks like your call to println() is outside your loop.
ChotaDon for(int i=1;i<=10;i++) System.out.println(N+" x "+i+" = "+N*i);
chinmay_verneka1 for (int i=1; i<=10; i++) { res=N*i; System.out.println(" "+N+" x "+i+" = " +res); }
I got the same output but it say it is wrong
shahharsh299 public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); int h=2; for(int i=1;i<=10;i++)
{
System.out.println(N +" x " + i + " = " + N*i);} scanner.close(); }
shahharsh299 public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); int h=2; for(int i=1;i<=10;i++)
{
System.out.println(N +" x " + i + " = " + N*i);} scanner.close(); }
shahharsh299 LOL
krishnadulaldal1 public class Solution {
private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); int i; for(i=1;i<=10;i++){ System.out.println(N+" x "+i+" = "+(N*i));} scanner.close(); } }
amoghrm111 what is that scanner skip statement??please help me
patsingh1997 import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;
public class Solution {
private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?"); for(int i=1;i<=10;i++){ System.out.println(N+" x "+i+" = "+N*i); } scanner.close(); }} //But this one is right WFK
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