Counting multiples of 3 up to a given numberSPOJ “SBANK - Sorting Bank Accounts” TLEOptimal way to annihilate a list by removing items from the endsCounting number of inversionsCount the number of inversions using Binary Indexed Tree in an arrayTime Limit Exceeded for ETF - Euler Totient Function at SpojCounting pairs of integers whose sum is less than a given thresholdCodeChef PRIME1 problem in PythonOptimised Python Code gives TLE for PRIME1Count numbers with atleast one common factor, excluding oneSplit a given number so that their sum adds to another given number
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Counting multiples of 3 up to a given number
SPOJ “SBANK - Sorting Bank Accounts” TLEOptimal way to annihilate a list by removing items from the endsCounting number of inversionsCount the number of inversions using Binary Indexed Tree in an arrayTime Limit Exceeded for ETF - Euler Totient Function at SpojCounting pairs of integers whose sum is less than a given thresholdCodeChef PRIME1 problem in PythonOptimised Python Code gives TLE for PRIME1Count numbers with atleast one common factor, excluding oneSplit a given number so that their sum adds to another given number
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I'm attempting a practice problem in Codeforces that requires you to find the number of multiples of 3 less than or equal to a given number. I have written the following code but it gives a TLE once the test case reaches the order of 108.
int x = Reader.nextInt();
int count = 0;
for(int i = 3; i<=x; i++)
if(i%3 == 0)
count++;
System.out.println(count);
Is there any way I can further optimize this code? The test cases max out at 109.
java programming-challenge time-limit-exceeded
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
New contributor
therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I'm attempting a practice problem in Codeforces that requires you to find the number of multiples of 3 less than or equal to a given number. I have written the following code but it gives a TLE once the test case reaches the order of 108.
int x = Reader.nextInt();
int count = 0;
for(int i = 3; i<=x; i++)
if(i%3 == 0)
count++;
System.out.println(count);
Is there any way I can further optimize this code? The test cases max out at 109.
java programming-challenge time-limit-exceeded
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
New contributor
therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
2
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago
add a comment |
$begingroup$
I'm attempting a practice problem in Codeforces that requires you to find the number of multiples of 3 less than or equal to a given number. I have written the following code but it gives a TLE once the test case reaches the order of 108.
int x = Reader.nextInt();
int count = 0;
for(int i = 3; i<=x; i++)
if(i%3 == 0)
count++;
System.out.println(count);
Is there any way I can further optimize this code? The test cases max out at 109.
java programming-challenge time-limit-exceeded
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
New contributor
therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm attempting a practice problem in Codeforces that requires you to find the number of multiples of 3 less than or equal to a given number. I have written the following code but it gives a TLE once the test case reaches the order of 108.
int x = Reader.nextInt();
int count = 0;
for(int i = 3; i<=x; i++)
if(i%3 == 0)
count++;
System.out.println(count);
Is there any way I can further optimize this code? The test cases max out at 109.
java programming-challenge time-limit-exceeded
java programming-challenge time-limit-exceeded
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
New contributor
therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
New contributor
therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
edited 5 hours ago
mdfst13
18k6 gold badges23 silver badges58 bronze badges
18k6 gold badges23 silver badges58 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
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asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
asked 12 hours ago
therealshankmantherealshankman
285 bronze badges
285 bronze badges
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therealshankman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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Check out our Code of Conduct.
3
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
2
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago
add a comment |
3
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
2
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago
3
3
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
2
2
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you are insisting on counting, instead of computing the answer, you can optimize your counting loop by just looping over the multiples of 3:
for(int i = 3; i <= x; i += 3)
count++;
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
$endgroup$
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of thefor()statement is proper, and can improve readability of the code.
$endgroup$
– AJNeufeld
9 hours ago
|
show 2 more comments
$begingroup$
Use your brain, not brute force. How would you solve the problem if it were presented in a math test, and you couldn't use a computer?
Your solution requires $O(x)$ time; it should be possible in $O(1)$.
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
$endgroup$
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
add a comment |
$begingroup$
If I were to compute the answer, I'd do (language shown is JavaScript):
var n = 1e8;
var v = Math.floor(n / 3);
6 has two multiples of three less than or equal to it: 6 / 3 == 2
127 has 42: 127 / 3 == 42.3333, thus Math.floor(127 / 3) == 42.
Likewise, 3 * n will have n multiples of three, including three itself.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
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VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you are insisting on counting, instead of computing the answer, you can optimize your counting loop by just looping over the multiples of 3:
for(int i = 3; i <= x; i += 3)
count++;
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
$endgroup$
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of thefor()statement is proper, and can improve readability of the code.
$endgroup$
– AJNeufeld
9 hours ago
|
show 2 more comments
$begingroup$
If you are insisting on counting, instead of computing the answer, you can optimize your counting loop by just looping over the multiples of 3:
for(int i = 3; i <= x; i += 3)
count++;
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
$endgroup$
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of thefor()statement is proper, and can improve readability of the code.
$endgroup$
– AJNeufeld
9 hours ago
|
show 2 more comments
$begingroup$
If you are insisting on counting, instead of computing the answer, you can optimize your counting loop by just looping over the multiples of 3:
for(int i = 3; i <= x; i += 3)
count++;
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
$endgroup$
If you are insisting on counting, instead of computing the answer, you can optimize your counting loop by just looping over the multiples of 3:
for(int i = 3; i <= x; i += 3)
count++;
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
answered 11 hours ago
AJNeufeldAJNeufeld
10.2k1 gold badge8 silver badges33 bronze badges
10.2k1 gold badge8 silver badges33 bronze badges
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of thefor()statement is proper, and can improve readability of the code.
$endgroup$
– AJNeufeld
9 hours ago
|
show 2 more comments
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of thefor()statement is proper, and can improve readability of the code.
$endgroup$
– AJNeufeld
9 hours ago
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
you missed a trick here, appending count++ in the loop
$endgroup$
– dfhwze
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
Yeah this works. I needed to count it as it was mentioned explicitly in the question. But I hadn't even figured the simple computation until a few minutes ago.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
$begingroup$
@dfhwze how would that work?
$endgroup$
– therealshankman
11 hours ago
1
1
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
$begingroup$
for(int i = 3; i <= x; i += 3, count++);
$endgroup$
– dfhwze
11 hours ago
1
1
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of the
for() statement is proper, and can improve readability of the code.$endgroup$
– AJNeufeld
9 hours ago
$begingroup$
@dfhwze That “trick” doesn’t make the code run any faster, and can lead to errors if you forget/don’t notice the trailing semicolon. I would call it out reviewing other people’s code here on Code Review. I certainly didn’t “miss” it, and highly recommend against it. When iterating over a non-trivial body, with two or more iterators, incrementing all of them in the increment portion of the
for() statement is proper, and can improve readability of the code.$endgroup$
– AJNeufeld
9 hours ago
|
show 2 more comments
$begingroup$
Use your brain, not brute force. How would you solve the problem if it were presented in a math test, and you couldn't use a computer?
Your solution requires $O(x)$ time; it should be possible in $O(1)$.
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
$endgroup$
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
add a comment |
$begingroup$
Use your brain, not brute force. How would you solve the problem if it were presented in a math test, and you couldn't use a computer?
Your solution requires $O(x)$ time; it should be possible in $O(1)$.
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
$endgroup$
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
add a comment |
$begingroup$
Use your brain, not brute force. How would you solve the problem if it were presented in a math test, and you couldn't use a computer?
Your solution requires $O(x)$ time; it should be possible in $O(1)$.
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
$endgroup$
Use your brain, not brute force. How would you solve the problem if it were presented in a math test, and you couldn't use a computer?
Your solution requires $O(x)$ time; it should be possible in $O(1)$.
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
edited 6 hours ago
dfhwze
5,2551 gold badge7 silver badges36 bronze badges
5,2551 gold badge7 silver badges36 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
answered 11 hours ago
200_success200_success
135k21 gold badges172 silver badges441 bronze badges
135k21 gold badges172 silver badges441 bronze badges
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
add a comment |
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
1
1
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
$begingroup$
The title does state counting, not computing :p
$endgroup$
– dfhwze
11 hours ago
1
1
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
$begingroup$
Yep, I figured. I'll keep the question up as a reminder to me that most simple problems will have simple solutions.
$endgroup$
– therealshankman
11 hours ago
1
1
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
$begingroup$
@dfhwze That's only since someone butchered the title. It originally said finding.
$endgroup$
– mdfst13
5 hours ago
add a comment |
$begingroup$
If I were to compute the answer, I'd do (language shown is JavaScript):
var n = 1e8;
var v = Math.floor(n / 3);
6 has two multiples of three less than or equal to it: 6 / 3 == 2
127 has 42: 127 / 3 == 42.3333, thus Math.floor(127 / 3) == 42.
Likewise, 3 * n will have n multiples of three, including three itself.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
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VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
add a comment |
$begingroup$
If I were to compute the answer, I'd do (language shown is JavaScript):
var n = 1e8;
var v = Math.floor(n / 3);
6 has two multiples of three less than or equal to it: 6 / 3 == 2
127 has 42: 127 / 3 == 42.3333, thus Math.floor(127 / 3) == 42.
Likewise, 3 * n will have n multiples of three, including three itself.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
add a comment |
$begingroup$
If I were to compute the answer, I'd do (language shown is JavaScript):
var n = 1e8;
var v = Math.floor(n / 3);
6 has two multiples of three less than or equal to it: 6 / 3 == 2
127 has 42: 127 / 3 == 42.3333, thus Math.floor(127 / 3) == 42.
Likewise, 3 * n will have n multiples of three, including three itself.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If I were to compute the answer, I'd do (language shown is JavaScript):
var n = 1e8;
var v = Math.floor(n / 3);
6 has two multiples of three less than or equal to it: 6 / 3 == 2
127 has 42: 127 / 3 == 42.3333, thus Math.floor(127 / 3) == 42.
Likewise, 3 * n will have n multiples of three, including three itself.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
answered 41 mins ago
VisualPlugin RōbloxVisualPlugin Rōblox
1
1
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
VisualPlugin Rōblox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
add a comment |
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
$begingroup$
I'm using Java here. Doing long/int division in Java ignores the decimal section anyways. But thanks for the answer, I'll keep it in mind for a later use.
$endgroup$
– therealshankman
13 mins ago
add a comment |
therealshankman is a new contributor. Be nice, and check out our Code of Conduct.
therealshankman is a new contributor. Be nice, and check out our Code of Conduct.
therealshankman is a new contributor. Be nice, and check out our Code of Conduct.
therealshankman is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
Can you think of a simple mathematical formula which computes the result directly, without the need for a loop?
$endgroup$
– Martin R
12 hours ago
2
$begingroup$
Is it dividing the given number by 3? Oh no
$endgroup$
– therealshankman
12 hours ago