How to check if string is entirely made of same substring?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}

I have to create a function which takes a string should return true or false based on the string. The length of given string is always greater than 1.

"aa" // true(entirely contains two strings "a")

"aaa" //true(entirely contains three string "a")

"abcabcabc" //true(entirely containas three strings "abc")



"aba" //false(At least there should be two same substrings and nothing more)

"ababa" //false("ab" exists twice but "a" is extra so false)

I have created a function but its

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

Checking of this is part of real problem. I can't afford a non-efficient solution like this. First of all its looping through half of the string. Second problem is that its using replace() in each loop which makes it slow.

Can some please suggest be a better solution regarding performance?

asked 56 mins ago

Maheer Ali

12.6k1128

2

This link may be useful to you. I always find geekforgeeks as a good source for algorithm problems - geeksforgeeks.org/…

– Leron
46 mins ago

I think its a problem of checking whether a common substring occurs throughout uniformly

– Kunal Mukherjee
39 mins ago

add a comment |

I have to create a function which takes a string should return true or false based on the string. The length of given string is always greater than 1.

"aa" // true(entirely contains two strings "a")

"aaa" //true(entirely contains three string "a")

"abcabcabc" //true(entirely containas three strings "abc")



"aba" //false(At least there should be two same substrings and nothing more)

"ababa" //false("ab" exists twice but "a" is extra so false)

I have created a function but its

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

asked 56 mins ago

Maheer Ali

12.6k1128

2

This link may be useful to you. I always find geekforgeeks as a good source for algorithm problems - geeksforgeeks.org/…

– Leron
46 mins ago

I think its a problem of checking whether a common substring occurs throughout uniformly

– Kunal Mukherjee
39 mins ago

add a comment |

I have to create a function which takes a string should return true or false based on the string. The length of given string is always greater than 1.

"aa" // true(entirely contains two strings "a")

"aaa" //true(entirely contains three string "a")

"abcabcabc" //true(entirely containas three strings "abc")



"aba" //false(At least there should be two same substrings and nothing more)

"ababa" //false("ab" exists twice but "a" is extra so false)

I have created a function but its

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

asked 56 mins ago

Maheer Ali

12.6k1128

I have to create a function which takes a string should return true or false based on the string. The length of given string is always greater than 1.

"aa" // true(entirely contains two strings "a")

"aaa" //true(entirely contains three string "a")

"abcabcabc" //true(entirely containas three strings "abc")



"aba" //false(At least there should be two same substrings and nothing more)

"ababa" //false("ab" exists twice but "a" is extra so false)

I have created a function but its

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

function check(str){

  if(!(str.length && str.length - 1)) return false;

  let temp = '';

  for(let i = 0;i<=str.length/2;i++){

    temp += str[i] 

    //console.log(str.replace(new RegExp(temp,"g"),''))

    if(!str.replace(new RegExp(temp,"g"),'')) return true;

  }

  return false;

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

javascript algorithm

asked 56 mins ago

Maheer Ali

12.6k1128

asked 56 mins ago

Maheer Ali

12.6k1128

asked 56 mins ago

Maheer Ali

12.6k1128

asked 56 mins ago

Maheer Ali

12.6k1128

asked 56 mins ago

Maheer Ali

12.6k1128

2

This link may be useful to you. I always find geekforgeeks as a good source for algorithm problems - geeksforgeeks.org/…

– Leron
46 mins ago

I think its a problem of checking whether a common substring occurs throughout uniformly

– Kunal Mukherjee
39 mins ago

add a comment |

2

This link may be useful to you. I always find geekforgeeks as a good source for algorithm problems - geeksforgeeks.org/…

– Leron
46 mins ago

I think its a problem of checking whether a common substring occurs throughout uniformly

– Kunal Mukherjee
39 mins ago

This link may be useful to you. I always find geekforgeeks as a good source for algorithm problems - geeksforgeeks.org/…

– Leron
46 mins ago

I think its a problem of checking whether a common substring occurs throughout uniformly

– Kunal Mukherjee
39 mins ago

add a comment |

3 Answers
3

active

oldest

votes

You can do it by capturing group and backreference, just check it's the repetition of first captured value.

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

In the above RegExp :

^ and $ stands for start and end anchors to predict the position.

(.+) captures any pattern and captures the value(except n).

1 is backreference of first captured value and 1+ would check for repetition of captured value.

Regex explanation here

Performance difference : https://jsperf.com/reegx-and-loop/1

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

5

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave: If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n). but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

|
show 8 more comments

The fastest algorithmic approach is building Z-function in linear time

C++ implementation for reference (I can't say what JS implementation is good)

vector<int> z_function(string s) {

    int n = (int) s.length();

    vector<int> z(n);

    for (int i = 1, l = 0, r = 0; i < n; ++i) {

        if (i <= r)

            z[i] = min (r - i + 1, z[i - l]);

        while (i + z[i] < n && s[z[i]] == s[i + z[i]])

            ++z[i];

        if (i + z[i] - 1 > r)

            l = i, r = i + z[i] - 1;

    }

    return z;

}

Then you need to check indexes i that divide n.

If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true

answered 23 mins ago

MBo

50.6k23152

add a comment |

-1

One of the simple idea, replace the string with the substring of "" and it any text exist then it is false ,else it is true.

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

answered 46 mins ago

Vinod kumar G

435412

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55823298%2fhow-to-check-if-string-is-entirely-made-of-same-substring%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can do it by capturing group and backreference, just check it's the repetition of first captured value.

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

In the above RegExp :

^ and $ stands for start and end anchors to predict the position.

(.+) captures any pattern and captures the value(except n).

1 is backreference of first captured value and 1+ would check for repetition of captured value.

Regex explanation here

Performance difference : https://jsperf.com/reegx-and-loop/1

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

5

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave: If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n). but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

|
show 8 more comments

You can do it by capturing group and backreference, just check it's the repetition of first captured value.

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

In the above RegExp :

^ and $ stands for start and end anchors to predict the position.

(.+) captures any pattern and captures the value(except n).

1 is backreference of first captured value and 1+ would check for repetition of captured value.

Regex explanation here

Performance difference : https://jsperf.com/reegx-and-loop/1

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

5

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave: If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n). but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

|
show 8 more comments

You can do it by capturing group and backreference, just check it's the repetition of first captured value.

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

In the above RegExp :

^ and $ stands for start and end anchors to predict the position.

(.+) captures any pattern and captures the value(except n).

1 is backreference of first captured value and 1+ would check for repetition of captured value.

Regex explanation here

Performance difference : https://jsperf.com/reegx-and-loop/1

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

You can do it by capturing group and backreference, just check it's the repetition of first captured value.

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

In the above RegExp :

^ and $ stands for start and end anchors to predict the position.

(.+) captures any pattern and captures the value(except n).

1 is backreference of first captured value and 1+ would check for repetition of captured value.

Regex explanation here

Performance difference : https://jsperf.com/reegx-and-loop/1

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

function check(str) {

  return /^(.+)1+$/.test(str)

}



console.log(check('aa')) //true

console.log(check('aaa')) //true 

console.log(check('abcabcabc')) //true

console.log(check('aba')) //false

console.log(check('ababa')) //false

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

edited 25 mins ago

answered 48 mins ago

Pranav C Balan

91k1491118

answered 48 mins ago

Pranav C Balan

91k1491118

answered 48 mins ago

Pranav C Balan

91k1491118

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

5

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave: If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n). but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

|
show 8 more comments

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

5

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave: If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n). but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

Can you explain to us what this line is doing return /^(.+)1+$/.test(str)

– Thanveer Shah
47 mins ago

Also what is the complexity of this solution? I'm not absolutely sure but it doesn't seem to be much faster than the one the OP has.

– Leron
43 mins ago

@PranavCBalan I'm not good at algorithms, that's why I write in the comments section. However I have several things to mention - the OP already has a working solution so he is asking for one that will give him better performance and you haven't explained how your solution will outperform his. Shorter doesn't mean faster. Also, from the link you gave:

If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n).

but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

If you use normal (TCS:no backreference, concatenation,alternation,Kleene star) regexp and regexp is already compiled then it's O(n).

but as you wrote you are using backreference so is it still O(n)?

– Leron
33 mins ago

@PranavCBalan Its showing both are fastest.

– Maheer Ali
28 mins ago

@MaheerAli : for me, it's showing existing code is 44% slower

– Pranav C Balan
27 mins ago

|
show 8 more comments

The fastest algorithmic approach is building Z-function in linear time

C++ implementation for reference (I can't say what JS implementation is good)

vector<int> z_function(string s) {

    int n = (int) s.length();

    vector<int> z(n);

    for (int i = 1, l = 0, r = 0; i < n; ++i) {

        if (i <= r)

            z[i] = min (r - i + 1, z[i - l]);

        while (i + z[i] < n && s[z[i]] == s[i + z[i]])

            ++z[i];

        if (i + z[i] - 1 > r)

            l = i, r = i + z[i] - 1;

    }

    return z;

}

Then you need to check indexes i that divide n.

If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true

answered 23 mins ago

MBo

50.6k23152

add a comment |

The fastest algorithmic approach is building Z-function in linear time

C++ implementation for reference (I can't say what JS implementation is good)

vector<int> z_function(string s) {

    int n = (int) s.length();

    vector<int> z(n);

    for (int i = 1, l = 0, r = 0; i < n; ++i) {

        if (i <= r)

            z[i] = min (r - i + 1, z[i - l]);

        while (i + z[i] < n && s[z[i]] == s[i + z[i]])

            ++z[i];

        if (i + z[i] - 1 > r)

            l = i, r = i + z[i] - 1;

    }

    return z;

}

Then you need to check indexes i that divide n.

If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true

answered 23 mins ago

MBo

50.6k23152

add a comment |

The fastest algorithmic approach is building Z-function in linear time

C++ implementation for reference (I can't say what JS implementation is good)

vector<int> z_function(string s) {

    int n = (int) s.length();

    vector<int> z(n);

    for (int i = 1, l = 0, r = 0; i < n; ++i) {

        if (i <= r)

            z[i] = min (r - i + 1, z[i - l]);

        while (i + z[i] < n && s[z[i]] == s[i + z[i]])

            ++z[i];

        if (i + z[i] - 1 > r)

            l = i, r = i + z[i] - 1;

    }

    return z;

}

Then you need to check indexes i that divide n.

If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true

answered 23 mins ago

MBo

50.6k23152

The fastest algorithmic approach is building Z-function in linear time

C++ implementation for reference (I can't say what JS implementation is good)

vector<int> z_function(string s) {

    int n = (int) s.length();

    vector<int> z(n);

    for (int i = 1, l = 0, r = 0; i < n; ++i) {

        if (i <= r)

            z[i] = min (r - i + 1, z[i - l]);

        while (i + z[i] < n && s[z[i]] == s[i + z[i]])

            ++z[i];

        if (i + z[i] - 1 > r)

            l = i, r = i + z[i] - 1;

    }

    return z;

}

Then you need to check indexes i that divide n.

If you find such i that i+z[i]=n then the string s can be compressed to the length i and you can return true

answered 23 mins ago

MBo

50.6k23152

answered 23 mins ago

MBo

50.6k23152

answered 23 mins ago

MBo

50.6k23152

answered 23 mins ago

MBo

50.6k23152

add a comment |

-1

One of the simple idea, replace the string with the substring of "" and it any text exist then it is false ,else it is true.

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

answered 46 mins ago

Vinod kumar G

435412

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

add a comment |

-1

One of the simple idea, replace the string with the substring of "" and it any text exist then it is false ,else it is true.

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

answered 46 mins ago

Vinod kumar G

435412

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

add a comment |

-1

One of the simple idea, replace the string with the substring of "" and it any text exist then it is false ,else it is true.

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

answered 46 mins ago

Vinod kumar G

435412

One of the simple idea, replace the string with the substring of "" and it any text exist then it is false ,else it is true.

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

'ababababa'.replace(/ab/gi,'')

"a" // return false

'abababab'.replace(/ab/gi,'')

 ""// return true

answered 46 mins ago

Vinod kumar G

435412

answered 46 mins ago

Vinod kumar G

435412

answered 46 mins ago

Vinod kumar G

435412

answered 46 mins ago

Vinod kumar G

435412

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

add a comment |

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

What about abcabcabc or unicornunicorn?

– adiga
44 mins ago

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Gfyuki