int main(){
int input;
int bin = 0, i = 1;
print("Please input a number");
input = num.parse(stdin.readLineSync());
while(input > 0)
{
bin = bin + (input % 2)*i;
input = input/2;
i = i * 10;
}
return 0;
}
It returned infinite numbers.
You just need to take care of double to int conversion: input = (input/2).floor()
See this working code:
void main() {
int input;
int bin = 0, i = 1;
input = 5;
while(input > 0)
{
bin = bin + (input % 2)*i;
input = (input/2).floor();
i = i * 10;
}
print(bin);
}
Here is a version of the above function that:
uses the integer division
with a ternary conditional operator, avoids the conversion to string.
sets the most significant bit to the left (bin = (dec % 2) + bin; which is what most people expects, but is not what the original snippet did)
String dec2bin(int dec) {
var bin = '';
while (dec > 0) {
bin = (dec % 2 == 0 ? '0' : '1') + bin;
dec ~/= 2;
}
return bin;
}
P.S: But, of course, one can simply write:
var bin = num.toRadixString(2);
There is no real need to write your own dec2bin function.
As the int result can become easily big in length and the max int value in dart is 2e53 (and much less if you compile to web). it's better to change the approach and return as a String.
String dec2bin(int decimal) {
String bin = '';
while (decimal > 0) {
bin = bin + (decimal % 2).toString();
decimal = (decimal / 2).floor();
}
return bin;
}
print(dec2bin(132070242815));
Result: 1111111110011111111111111111110101111
//number is always in int
static String decimalToBinary(int number) {
return number.toRadixString(2);
}
//binary is always in string
static int binaryToDecimal(String binary) {
return int.parse(binary, radix: 2);
}
Related
May I know how to store the N prime numbers, which I got from for loop in an array in dart?
import 'dart:io';
void main() {
// print('enter a start number');
// int a = int.parse(stdin.readLineSync()!);
print('enter a number');
int b = int.parse(stdin.readLineSync());
print('this are prime numbers');
primenum(b);
var z = '';
}
primenum(b) {
String string = "";
int a = 2;
outerLoop:
for (int i = a; i <= b; i++) {
for (int x = 2; x <= i / a; x++) {
if (i % x == 0) {
continue outerLoop;
}
}
var z = i.toString();
// print(z);
var h = z;
// String str = '';
string = string + h;
}
List d = string.split('');
print(d);
}
Using the above code, I am able to get those numbers in List. But the double-digit numbers are splitting.
May I know How to solve the above task? using dart.
The way you're doing string.split is splitting the string into a list of each individual character. Instead, you can add each prime number to a List directly without doing string manipulation.
primenum(b) {
List<String> d;
int a = 2;
outerLoop:
for (int i = a; i <= b; i++) {
for (int x = 2; x <= i / a; x++) {
if (i % x == 0) {
continue outerLoop;
}
}
d.add(i.toString());
}
print(d);
}
I created a function to calculate the selling price of an item. Each year, the price of the item will decrease by 3/4 of its original price. The problem with my function is it doesn't want to read the year variable regardless of its value. My function always returns 60000000. Can someone please tell me what's wrong with it?
int add(double year, double price) {
int i = 0;
while (i < year) {
double final_price = price * 3 / 4;
i++;
return final_price.round();
}
}
void main(List<String> arguments) {
double x = 3;
double y = 80000000;
int result = add(x, y);
print(result);
}
int add(double year, double price) {
int i = 0;
double final_price=price; // change 1
while (i < year) {
final_price = final_price* 3 / 4; // change 2
i++;
}
return final_price.round(); // change 3
}
void main(List<String> arguments) {
double x = 3;
double y = 80000000;
int result = add(x, y);
print(result);
}
Here you go. Returning a value from inside the while loop will stop the function execution on the first traversal only.
Also you need to make final price equal to price because final_price's value is not being changed as price remains same and i increases.
what actually are you to trying to do ?
I fixed your code
int add(double year, double price) {
int i = 0;
double final_price = 0 ;
while (i < year) {
double f = price * 3 / 4;
i++;
final_price = f ;
}
return final_price.round();
}
void main() // you try to pass an arguments that never used I remove it
{
double x = 3;
double y = 800000;
int result = add(x, y);
print(result);
}
I am trying to solve 445. Add Two Numbers II from LeetCode where it is asked:
Given two non-empty linked lists representing two non-negative integers, add the two numbers and return it as a linked list. The most significant digit comes first and each of their nodes contain a single digit.
For some test cases, I am getting negative number from the sum method that I have implemented. I think it is impossible to get any negative digit in my code. Can you help me finding the bug?
Below is the code that I tried:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//ListNode previous = null;
int len1 = 0;
int len2 = 0;
ListNode head1 = l1;
ListNode head2 = l2;
while(l1!=null){
len1++;
l1=l1.next;
}
while(l2!=null){
len2++;
l2=l2.next;
}
int sum=0;
if (len1 >= len2){
sum= sum(head1,head2,len1,len2);
}else{
sum= sum(head2,head1,len2,len1);
}
String sumString = "" + sum;
ListNode extraHead = new ListNode(1);
ListNode copy = extraHead;
for(int i = 0;i<sumString.length();i++){
ListNode bit = new ListNode(Integer.parseInt(String.valueOf(sumString.charAt(i))));//Integer.parseInt(String.valueOf(sumString.charAt(i)))
extraHead.next = bit;
extraHead = extraHead.next;
}
return copy.next;
}
public int sum(ListNode l1, ListNode l2, int len1, int len2){
int diff = 0;
int resLen = 0;
diff = len1 - len2;
resLen = len1;
int[] res = new int[resLen];
ListNode fast = l1;
ListNode slow = l2;
for(int count = 0; count<diff; count++){
res[count] = fast.val;
fast = fast.next;
}
for(int count = diff;count < res.length;count++){
res[count] = fast.val + slow.val;
fast=fast.next;
slow=slow.next;
}
int sum = 0;
for(int i = len1;i>0;i-- ){
sum = sum + res[len1-i] * (int)Math.pow(10,i-1);
}
return sum;
}
}
Here is the error message that I get:
Error Details
java.lang.NumberFormatException: For input string: "-"
at line 68, java.base/java.lang.NumberFormatException.forInputString
at line 648, java.base/java.lang.Integer.parseInt
at line 776, java.base/java.lang.Integer.parseInt
at line 41, Solution.addTwoNumbers
at line 54, __DriverSolution__.__helper__
at line 87, __Driver__.main
Here is the input causing the error:
[3,9,9,9,9,9,9,9,9,9]
[7]
I want to generate a random six-digit number. I tried to use the Random class, but new Random().nextInt(999999) generates some numbers with less than six digits.
So you want just the numbers 100000 to (and including) 999999.
you can get a random number in this range (900000) and add 100000 to the random number you get:
var rng = new Random();
var code = rng.nextInt(900000) + 100000;
This will always give you a random number with 6 digits.
void main() {
var rnd = new math.Random();
var next = rnd.nextDouble() * 1000000;
while (next < 100000) {
next *= 10;
}
print(next.toInt());
}
you can also generate 6 different numbers and then concatenate them in one string and convert it to integer if you want
import 'dart:math';
main(){
var rndnumber="";
var rnd= new Random();
for (var i = 0; i < 6; i++) {
rndnumber = rndnumber + rnd.nextInt(9).toString();
}
print(rndnumber);
}
Here is a Dart extension method that will generate a non-negative random integer with a specified number of digits:
extension RandomOfDigits on Random {
/// Generates a non-negative random integer with a specified number of digits.
///
/// Supports [digitCount] values between 1 and 9 inclusive.
int nextIntOfDigits(int digitCount) {
assert(1 <= digitCount && digitCount <= 9);
int min = digitCount == 1 ? 0 : pow(10, digitCount - 1);
int max = pow(10, digitCount);
return min + nextInt(max - min);
}
}
In your case use it like this:
final random = Random();
print(random.nextIntOfDigits(6));
The following class will generate an integer with 'n' digits, or a string with 'n' digits.
The numeric method will be much faster, but is limited in the number of digits.
import 'dart:math';
class RandomDigits {
static const MaxNumericDigits = 17;
static final _random = Random();
static int getInteger(int digitCount) {
if (digitCount > MaxNumericDigits || digitCount < 1) throw new RangeError.range(0, 1, MaxNumericDigits, "Digit Count");
var digit = _random.nextInt(9) + 1; // first digit must not be a zero
int n = digit;
for (var i = 0; i < digitCount - 1; i++) {
digit = _random.nextInt(10);
n *= 10;
n += digit;
}
return n;
}
static String getString(int digitCount) {
String s = "";
for (var i = 0; i < digitCount; i++) {
s += _random.nextInt(10).toString();
}
return s;
}
}
void main() {
print(RandomDigits.getInteger(6));
print(RandomDigits.getString(36));
}
Output:
995723
198815207332880163668637448423456900
If you want to get 6 digit value from 0 to 999999, you can add leading 0 if the number is less than 6 digits.
String r = Random().nextInt(999999).toString().padLeft(6, '0');
// example output: 025328
i had the same problem.although there are 3-4 ways to tackle it but i find the below one simple and sorted.
simply check for the number of length match. find below code
Integer otp = new Random().nextInt(999999);
int noOfOtpDigit=6;
while(Integer.toString(otp).length()!=noOfOtpDigit) {
otp = new Random().nextInt(999999);
}
String otpString = String.valueOf(otp);
import 'dart:math';
void main() {
print(get6DigitNumber());
}
String get6DigitNumber(){
Random random = Random();
String number = '';
for(int i = 0; i < 6; i++){
number = number + random.nextInt(9).toString();
}
return number;
}
Here is your unlimited supply of six-digit random numbers
String getRandomNumber(){
final r = Random();
return List<int>.generate(6, (index) => r.nextInt(10)).fold<String>("", (prev, i) => prev += i.toString());
}
This is working for me in C#.
Random random = new Random();
string elementIndex = random.Next(100000, 999999).ToString();
I need to calculate the factorial of a decimal number, say 6.4, on iOS. I tried
double myFactorial = gamma(6.4);
But get the error "'gamma is unavailable': not avaiable on iOS". Is there a way to add the gamma function into iOS?
Have you tried:
tgamma(6.4)
I see it working in my code.
There's also:
double tgamma (double x)
float tgammaf (float x)
long double tgammal (long double x)
you can try like this logic may be this will well you.
- (int)factorial:(int)operand
{
if`enter code here`(operand < 0)
return -1;
else if (operand > 1)
return operand * [self factorial:operand - 1];
else
return 1;
}
and then
- (double)factorial:(double)operand
{
double output = operand;
if (output == 0) output = 1; // factorial of 0 is 1
else if (output < 0) output = NAN;
else if (output > 0)
{
if (fmod(output, floor(output)) == 0) // integer
output = round(exp(lgamma(output + 1)));
else // natural number
output = exp(lgamma(output + 1));
}
return output;
}
- (double)n:(double)n chooseR:(double)r
{
return round(exp((lgamma(n+1)) - (lgamma(r+1) + lgamma(n-r+1))));
}
- (double)n:(double)n pickR:(double)r
{
return round(exp(lgamma(n+1) - lgamma(n-r+1)));
}