I want to find the sum of the digits of the number entered in Flutter. I want to encode this algorithm.
for example
x=1992
result=1+9+9+2=21
how can i do this with flutter
You can do in this way.
import 'dart:io';
void main() {
print('Enter X');
int X = int.parse(stdin.readLineSync()!);
int result = 0;
for (int i = X; i > 0; i = (i / 10).floor()) {
result += (i % 10);
}
print('Sum of digits\n$result');
}
Output
Enter X
123456
Sum of digits
21
transform the number into an String using String stringValue = x.toString();
create an array from each char using List<String> result = stringValue.split('');
sum each number transforming back using int.parse(result)
void main(){
int x = 1992;
String stringValue = x.toString();
List<String> result = stringValue.split('');
int sum = 0;
for(int i = 0 ; i < result.length; i++) {
int value = int.parse(result[i]);
sum = sum + value;
}
print(sum);
}
Result: 21
Related
Write a program that prints all numbers from 1 to 1000 in a large to small (descending) order.
Expected Output:
Each line will have a total of five (5) numbers and each number will be separated with a '\t' (Tab) character.
void main() {
final output = StringBuffer();
final columns = 5;
final maxInt = 1000;
var index =1;
final values = List.generate(maxInt, (index) => index+1).reversed;
while(index <= maxInt) {
output.write('${values.elementAt(index - 1)}\t');
if (index % columns == 0) {
output.writeln();
}
index++;
}
print(output);
}
Her you go.
void main() {
for (int i = 1000; i >= 1; i-=5) {
print('${i} ${i-1} ${i-2} ${i-3} ${i-4}');
}
}
Is it possible to concatenate integers without converting to String first?
int _test1 = 123;
int _test2 = 456;
print(int.parse(("$_test1"+"$_test2"))); // 123456
you can do it like this
void main() {
int _test1 = 123;
int _test2 = 456;
int pow = 10;
while(_test2 >= pow)
pow *= 10;
print(_test1 *pow + _test2);
}
source : How to concatenate two integers in C
I’m dealing with BigDecimal in Java and I need to make 2 check against BigDecimal fields in my DTO:
Number of digits of full part (before point) < 15
Total number of
digits < 32 including scale (zeros after point)
What is the best way to implement it? I extremely don’t want toBigInteger().toString() and .toString()
I think this will work.
BigDecimal d = new BigDecimal("921229392299229.2922929292920000");
int fractionCount = d.scale();
System.out.println(fractionCount);
int wholeCount = (int) (Math.ceil(Math.log10(d.longValue())));
System.out.println(wholeCount);
I did some testing of the above method vs using indexOf and subtracting lengths of strings. The above seems to be signficantly faster if my testing methodology is reasonable. Here is how I tested it.
Random r = new Random(29);
int nRuns = 1_000_000;
// create a list of 1 million BigDecimals
List<BigDecimal> testData = new ArrayList<>();
for (int j = 0; j < nRuns; j++) {
String wholePart = r.ints(r.nextInt(15) + 1, 0, 10).mapToObj(
String::valueOf).collect(Collectors.joining());
String fractionalPart = r.ints(r.nextInt(31) + 1, 0, 10).mapToObj(
String::valueOf).collect(Collectors.joining());
BigDecimal d = new BigDecimal(wholePart + "." + fractionalPart);
testData.add(d);
}
long start = System.nanoTime();
// Using math
for (BigDecimal d : testData) {
int fractionCount = d.scale();
int wholeCount = (int) (Math.ceil(Math.log10(d.longValue())));
}
long time = System.nanoTime() - start;
System.out.println(time / 1_000_000.);
start = System.nanoTime();
//Using strings
for (BigDecimal d : testData) {
String sd = d.toPlainString();
int n = sd.indexOf(".");
int m = sd.length() - n - 1;
}
time = System.nanoTime() - start;
System.out.println(time / 1_000_000.);
}
Hi i have been wondering if there is a way in which to convert binary numbers into decimal fractions.
I know how to change base as an example through this code
String binary = "11110010";
//I'd like to change this line so it produces a decimal value
String denary = int.parse(binary, radix: 2).toRadixString(10);
If anyone still wondering how to convert decimal to binary and the inverse:
print(55.toRadixString(2)); // Outputs 110111
print(int.parse("110111", radix: 2)); Outputs 55
int binaryToDecimal(int n)
{
int num = n;
int dec_value = 0;
// Initializing base value to 1, i.e 2^0
int base = 1;
int temp = num;
while (temp) {
int last_digit = temp % 10;
temp = temp / 10;
dec_value += last_digit * base;
base = base * 2;
}
return dec_value;
}
int main()
{
int num = 10101001;
cout << binaryToDecimal(num) << endl;
}
This is my c++ solution but you can implement any language
I want to roundoff my number to greater value. for example if i have 234 i want to make it 300 and 4436 to 5000. I have tried it but i can roundoff my value not in hundred and thousants but in tens. like i have a value 7771 it roundoff to 7900 or 7800 but i want it in 8000.
int temp = lroundf([[arrayPercentage objectAtIndex:i]floatValue]);
maxPer = (((temp + 10)/10))*10;
How about this:
-(void)roundUpNumber:(NSInteger) num {
NSInteger numLength = [[NSString stringWithFormat:#"%ld",num] length];
NSInteger newNum = ceil(num/pow(10,numLength-1)) * pow(10,numLength -1);
NSLog(#"%ld",newNum);
}
int num, count = 0;
int originalNumber = 7771;
num = originalNumber;
while (num) {
num = num/10;
count ++;
}
int power = pow(10,(count -1));
int firstDigit = originalNumber / power;
int finalNumber = (firstDigit + 1)* power;
NSLog(#"final result : %d",finalNumber);