php while $count, want to add all the result - php
Im facing some issue on counting total number of output.
<?php
$count = 1;
while ($count <= 10)
{
echo "$count ";
++$count;
}
?>
Result output
1 2 3 4 5 6 7 8 9 10
so what i want is to add all the result that is
1+2+3+4+5+6+7+8+9+10 = ? in my same code?
Try
$count = 1;
$add=0;
while ($count <= 10)
{
$add=$add+$count;
echo "$count ";
++$count;
}
$count = 1;
$countall = 0;
while ($count <= 10)
{
echo "$count ";
$countall=$countall+$count;
$count++;
}
echo "$countall";
try this
Simply use the range function and array_sum to get the result
array_sum(range(1,10))
Evidently not what you are looking for, but if what you need is to calculate a summation, you can use this formula:
Using it you can calculate the result of adding all the values of $count in this code:
<?php
$count = 1;
while ($count <= $n)
{
echo $count.' ';
++$count;
}
?>
That will be:
<?php
$result = $n * ($n + 1) / 2;
?>
Which for $n = 10 is 55.
Related
Exam Qn: Convert do while loop to for loop (PHP)
Recently, my exams got over. My last exam was based on PHP. I got the following question for my exam:
"Convert the following script using for loop without affecting the output:-"
<?php
//Convert into for loop
$x = 0;
$count = 10;
do
{
echo ($count. "<BR>");
$count = $count - 2;
$x = $x + $count;
}
while($count < 1)
echo ($x);
?>
Please help me as my computer sir is out of station and I am really puzzled by it.
Well, If I understand well, You have to use "for" loop, instead of "do...while", but the printed text must not change.
Try:
$count = 10;
$x = 0;
$firstRun = true;
for(; $count > 1 || $firstRun;) {
$firstRun = false;
echo ($count . "<BR>");
$count -= 2;
$x = $x + $count;
}
echo ($x);
By the way loop is unnecessary, because $count will be greater than 1 after the first loop, so the while will get false.
EDIT
$firstRun to avoid infiniteLoop
$count in loop
EDIT
Fixed code for new requirement
Removed unnecessary code
Hmmm I don't know if your teacher wanted to own you... but the do{} will execute only once since $count is never < 1.
The output of your teacher's code is:
10
8
I presume there was a mistake in the code and the while would be while($count > 1) which would make more sense (since it's weird to ask for a loop to output only 10 8) and would result in this output:
10
8
6
4
2
20
Then a good for() loop would have been :
$x = 0;
$count = 10;
for($i = $count; $i > 1; $i -= 2)
{
$count -= 2;
echo $i . "<br>";
$x += $count;
}
echo $x;
Which will output the same values.
If you can, ask your teacher for this, and comment the answer ^^ ahahah
PHP nested while loop
I tried to do a while inside a while to print a multiplication table like,
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
But I got only 1, 2, 3, 4, 5.
Code:
$i = 1;
$x = 1;
while($i <= 5){
while($x <= 5){
echo $i * $x;
$x++;
}
echo "<br>";
$i++;
}
This is happening because you're not resetting $x when the inner loop completes its iteration. Try this instead:
$i = 1;
while($i <= 5) {
$x = 1;
while($x <= 5) {
echo $i * $x;
$x++;
}
echo "<br>";
$i++;
}
You need to reset $x, so:
$i = 1;
$x = 1;
while($i <= 5){
while($x <= 5){
echo $i * $x;
$x++;
}
$x = 1; // added this line
echo "<br>";
$i++;
}
Output:
12345
246810
3691215
48121620
510152025
You can then do what ever you want to format it.
More elabrate explanation:
First run:
It enters both outer and inner loops, showing the desired output for the first line. You end up with $i = 2 and $x = 6.
Second run:
Since $i is 2, it doesn't leave the outer loop, but $x is 6, so it doesn't enter the inner loop again.
Last* run:
It then keeps adding 1 to $i until it doesn't match the outer loop condition anymore and leaves you with that unwanted result.
Use this
This is because you have not initialized your $x after external while loop completes its one cycle. so after one cycle inner loops does not run
<?php
$i = 1;
while($i <= 5) {
$x = 1;
while($x <= 5) {
echo $i * $x;
$x++;
}
echo "<br>";
$i++;
}
DEMO ONLINE
php code:
$i = 1;
while($i <= 5){
$x = 1;
while($x <= 5){
echo $i * $x." ";
$x++;
}
echo "<br/>";
$i++;
}
result:
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
I am facing a small bump printing number pyramids , still a newbie to php and programming
What i want to print is
1
3 5
7 9 11
With my current code , that is ...
<?php
function Odd($limit='20'){
$c = 1;
while($c <= $limit){
if ($c % 2!=0){
echo $c ;
echo "<br/>";
}
$c++ ;
}
}
Print Odd();
?>
i am getting
1
3
5
7
9
11
Can someone please guide me the right way ?
Aaah ... ok.^^ Now i got it.
Its pretty easy: You need another variable which counts up and one which limits the breakposition. Looks like this:
<?php
function Odd($limit='40'){
$c = 1;
$count = 0;
$break = 1;
while($c <= $limit){
if ($c % 2!=0){
echo $c . " ";
$count++;
if($count === $break) {
echo "<br/>";
$break++;
$count = 0;
}
}
$c++ ;
}
}
Print Odd();
?>
Output till 40:
1
3 5
7 9 11
13 15 17 19
21 23 25 27 29
31 33 35 37 39
Edit: Code for your new request:
<?php
function Odd($limit='40'){
$c = 1;
$count = 0;
$break = 1;
while($c <= $limit){
echo $c . " ";
$count++;
if($count === $break) {
echo "<br/>";
$break++;
$count = 0;
}
$c++ ;
}
}
Print Odd();
?>
So if I understand correctly you want to output something like that:
1
3 5
7 9 11
13 15 17 19
Here is my solution:
function Odd($limit='20'){
$c = 1;$some_array = array();
while($c <= $limit){
if ($c % 2!=0){
$some_array[]=$c;
}
$c++ ;
}
return $some_array;
}
$array = Odd();
$nr =0;
$j=1;
foreach ($array as $key => $value) {
echo $value.' ';$nr++;
if($nr==$j){
echo '<br />';
$nr=0;
$j++;
}
}
Hope this helps!
From your question it Seems you are really new to programming so before writing any program first of all observe the question properly:
For example for the question above it is clear that is an triangle of odd numbers.
now the number of odd numbers on each row is equal to the row
i.e 1st row contains 1 number ,2nd contains 2 and it continues...
Now what we do is take an variable to count the no of rows say $row and the other will be $limit .
<?php
function odd($limit){
$row=1;
$current_number=1;
while($current_number<=$limit){
for($i=1;$i<=$row;$i++){
echo $current_number." ";
$current_number=$current_number+2;//incrementing numbers by 2 if you want to increment by 1 i.e print all numbers replace 2 by 1
}
$row++;
echo "<br/>";//for new line
}
}
To run above function you need to call it and pass the value of $limit.To do it just type anywhere outside of this function.
odd(20);
Watch this running here:
how to find the sum of all the multiples of 3 or 5 below 1000 in php, issue?
i have an small issue with the way this problem is resolved.
some would say: println((0 /: ((0 until 1000).filter(x => x % 3 == 0 || x % 5 == 0))) (_+_)) is the solution witch adds to 233168
my way was to do:
$maxnumber = 1000;
for ($i = 3; $i < $maxnumber; $i += 3)
{
$t += $i;
echo $i.',';
}
echo '<br>';
for ($j = 5; $j < $maxnumber; $j += 5)
{
$d += $j;
echo $j.',';
}
echo '<br>';
echo $t;
echo '<br>';
echo $d;
echo '<br>';
echo $t+$d;
this will give me :
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198,201,204,207,210,213,216,219,222,225,228,231,234,237,240,243,246,249,252,255,258,261,264,267,270,273,276,279,282,285,288,291,294,297,300,303,306,309,312,315,318,321,324,327,330,333,336,339,342,345,348,351,354,357,360,363,366,369,372,375,378,381,384,387,390,393,396,399,402,405,408,411,414,417,420,423,426,429,432,435,438,441,444,447,450,453,456,459,462,465,468,471,474,477,480,483,486,489,492,495,498,501,504,507,510,513,516,519,522,525,528,531,534,537,540,543,546,549,552,555,558,561,564,567,570,573,576,579,582,585,588,591,594,597,600,603,606,609,612,615,618,621,624,627,630,633,636,639,642,645,648,651,654,657,660,663,666,669,672,675,678,681,684,687,690,693,696,699,702,705,708,711,714,717,720,723,726,729,732,735,738,741,744,747,750,753,756,759,762,765,768,771,774,777,780,783,786,789,792,795,798,801,804,807,810,813,816,819,822,825,828,831,834,837,840,843,846,849,852,855,858,861,864,867,870,873,876,879,882,885,888,891,894,897,900,903,906,909,912,915,918,921,924,927,930,933,936,939,942,945,948,951,954,957,960,963,966,969,972,975,978,981,984,987,990,993,996,999
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245,250,255,260,265,270,275,280,285,290,295,300,305,310,315,320,325,330,335,340,345,350,355,360,365,370,375,380,385,390,395,400,405,410,415,420,425,430,435,440,445,450,455,460,465,470,475,480,485,490,495,500,505,510,515,520,525,530,535,540,545,550,555,560,565,570,575,580,585,590,595,600,605,610,615,620,625,630,635,640,645,650,655,660,665,670,675,680,685,690,695,700,705,710,715,720,725,730,735,740,745,750,755,760,765,770,775,780,785,790,795,800,805,810,815,820,825,830,835,840,845,850,855,860,865,870,875,880,885,890,895,900,905,910,915,920,925,930,935,940,945,950,955,960,965,970,975,980,985,990,995
$t - 166833
$d - 99500
and total:
266333
why am i wrong?
Some numbers are multiples of both 3 and 5. (Your algorithm adds these numbers to the total twice.)
Because 6 * 5 == 30 and 10 * 3 == 30, you're adding the some numbers up twice.
$sum = 0;
$i = 0;
foreach(range(0, 999) as $i) {
if($i % 3 == 0 || $i % 5 == 0) $sum += $i;
}
Because you double-count numbers that are multiple of both 3 and 5, i.e. multiples of 15.
You can account for this naively by subtracting all multiples of 15.
for ($j = 15; $j < $maxnumber; $j += 15)
{
$e += $j;
echo $j.',';
}
$total = $total - $d;
In your case, if it is 15, you will add the number twice.
Try this:
$t = 0;
$d = 0;
for ($i = 0; $i <= $maxnumber; $i++){
if ($i % 3 == 0)
$t+= $i;
else if ($i % 5 == 0)
$d += $i;
}
echo $t.'<br>'.$d;
I think that in your code, if a number is a multiple of 3 and 5, it is added twice. Take 15 for example. It's in your list of multiples of 3 and in the list of multiples of 5. Is this the behaviour you want?
One of the best approach to this solution (to achieve optimum time complexity), run an Arithmetic Progression series and find the number of terms in all series by using AP formula: T=a+(n-1)d, then find sum by : S=n/2[2*a+(n-1)d]
where : a=first term ,n=no. of term , d=common deference, T=nth term
The code solution below has been implemented to suit the question above - so the values 3 and 5 are hard-coded. However, the function can modified such that values are passed in as variable parameters.
function solution($number){
$val1 = 3;
$val2 = 5;
$common_term = $val1 * $val2;
$sum_of_terms1 = calculateSumofMulitples($val1,$number);
$sum_of_terms2 = calculateSumofMulitples($val2,$number);
$sum_of_cterms = calculateSumofMulitples($common_term,$number);
$final_result = $sum_of_terms1 + $sum_of_terms2 - $sum_of_cterms;
return $final_result;
}
function calculateSumofMulitples($val, $number)
{
//first, we begin by running an aithmetic prograssion for $val up to $number say N to get the no of terms [using T=a +(n-1)d]
$no_of_terms = (int) ($number / $val);
if($number % $val == 0) $no_of_terms = (int) ( ($number-1)/$val ); //since we are computing for a no of terms below N and not up to/exactly/up to N. if N divides val with no remainder, use no of terms = (N-1)/val
//second, we compute the sum of terms [using Sn = n/2[2a + (n-1)d]
$sum_of_terms = ($no_of_terms * 0.5) * ( (2*$val) + ($no_of_terms - 1) * $val );
// sum of multiples
return $sum_of_terms;
}
You can run a single loop checking whether the number is multiple of 3 OR 5:
for ($i = 0; $i < $maxnumber; $i++)
{
if($i%3 || $i%5){
$t += $i;
echo $i.',';}
}
I think the original code is not including numbers which are multiples of both 3 and 5 in the total: if the test for multiple of 3 matches, it takes that and goes on. If you total the multiples of 15 up to 1000, you get 33165, which is exactly the difference between your total, 266333, and the original total, 233168.
Here's my solution to the question:
<?php
$sum = 0;
$arr = [];
for($i = 1; $i < 1000; $i++){
if((int)$i % 3 === 0 || (int)$i % 5 === 0)
{
$sum += $i;
array_push($arr,$i);
}
}
echo $sum;
echo '<br>';
print_r($arr);//Displays the values meeting the criteria as an array of values
Get the sum of all digits in a numeric string
How do I find the sum of all the digits in a number in PHP?
array_sum(str_split($number)); This assumes the number is positive (or, more accurately, that the conversion of $number into a string generates only digits).
Artefactos method is obviously unbeatable, but here an version how one could do it "manually":
$number = 1234567890;
$sum = 0;
do {
$sum += $number % 10;
}
while ($number = (int) ($number / 10));
This is actually faster than Artefactos method (at least for 1234567890), because it saves two function calls.
Another way, not so fast, not single line simple
<?php
$n = 123;
$nstr = $n . "";
$sum = 0;
for ($i = 0; $i < strlen($nstr); ++$i)
{
$sum += $nstr[$i];
}
echo $sum;
?>
It also assumes the number is positive.
function addDigits($num) {
if ($num % 9 == 0 && $num > 0) {
return 9;
} else {
return $num % 9;
}
}
only O(n)
at LeetCode submit result：
Runtime: 4 ms, faster than 92.86% of PHP online submissions for Add Digits.
Memory Usage: 14.3 MB, less than 100.00% of PHP online submissions for Add Digits.
<?php
// PHP program to calculate the sum of digits
function sum($num) {
$sum = 0;
for ($i = 0; $i < strlen($num); $i++){
$sum += $num[$i];
}
return $sum;
}
// Driver Code
$num = "925";
echo sum($num);
?>
Result will be 9+2+5 = 16
Try the following code:
<?php
$num = 525;
$sum = 0;
while ($num > 0)
{
$sum= $sum + ($num % 10);
$num= $num / 10;
}
echo "Summation=" . $sum;
?>
If interested with regex:
array_sum(preg_split("//", $number));
<?php
echo"----Sum of digit using php----";
echo"<br/ >";
$num=98765;
$sum=0;
$rem=0;
for($i=0;$i<=$num;$i++)
{
$rem=$num%10;
$sum=$sum+$rem;
$num=$num/10;
}
echo "The sum of digit 98765 is ".$sum;
?>
-----------------Output-------------
----Sum of digit using php----
The sum of digit 98765 is 35
// math before code
// base of digit sums is 9
// the product of all numbers multiplied by 9 equals 9 as digit sum
$nr = 58821.5712; // any number
// Initiallization
$d = array();
$d = explode(".",$nr); // cut decimal digits
$fl = strlen($d[1]); // count decimal digits
$pow = pow(10 ,$fl); // power up for integer
$nr = $nr * $pow; // make float become integer
// The Code
$ds = $nr % 9; // modulo of 9
if($ds == 0) $ds=9; // cancel out zeros
echo $ds;
Assume you want to find the sum of the digits of a number say 2395 the simplest solution would be to first split the digits and find out the sum then concatenate all the numbers into one single number. <?php $number=2; $number1=3; $number2=9; $number3=5; $combine=$number.$number1.$number2.$number3; $sum=$number+$number1+$number2+$number3; echo "The sum of $combine is $sum"; ?>
One way of getting sum of digit however this is a slowest route. $n=123; while(($n=$n-9)>9); echo "n: $n";
<html>
<head>
<title>detail</title>
</head>
<body>
<?php
$n = 123;
$sum=0; $n1=0;
for ($i =0; $i<=strlen($n);$i++)
{
$n1=$n%10;
$sum += $n1;
$n=$n/10;
}
echo $sum;
?>
</body>
</html>
Here's the code.. Please try this
<?php
$d=0;
$num=12345;
$temp=$num;
$sum=0;
while($temp>1)
{
$temp=$temp/10;
$d++;
}
echo "Digits Are : $d </br>";
for (;$num>1;)
{
$d=$num%10;
$num=$num/10;
$sum=$sum+$d;
}
echo "Sum of Digits is : $sum";
?>