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Guaguapi's blog

By Guaguapi, history, 4 weeks ago, In English,

Here is the problem 1221 F

Here is the code : Accepted ; Wrong answer ;

 
 
 
 
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4 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

it also give same wrong answer for C++11

your 'tot' operation look unbehaviour

it should be

X[tot + 1] = X[tot] + 1, tot++;

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    4 weeks ago, # ^ |
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    You mean that the operation ++tot doesn't same in C++17 and C++14 ? sorry for my poor English.

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      4 weeks ago, # ^ |
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      He means "undefined behavior". It's a kind of coding error and you shouldn't do that.

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    4 weeks ago, # ^ |
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    it also got accepted with C++ 11 ,here:61043518

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4 weeks ago, # |
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I could not find any operation for this behaviour.

Since it failed on 2nd test case so you can put debug statement with condition (n==5) so it won't fail on 1 case and you can know where it start changing behaviour.

If you find the mistake please update in post as well, i would like to know where it fail

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    4 weeks ago, # ^ |
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    See This answer in stack-overflow. In your submission X[++tot] = X[tot-1] + 1; is undefined behavior for c++ versions less than c++17: The left hand side of the = operator can be executed before or after the right hand side. In c++17, it's guaranteed that the right side will be executed before the left side. I think you assumed that in X[tot-1] tot's value was increased by 1 because of X[++tot], but actually tot is increasing later.

    Thanks again ShafinKhadem !

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4 weeks ago, # |
  Vote: I like it +18 Vote: I do not like it

See This answer in stack-overflow. In your submission X[++tot] = X[tot-1] + 1; is undefined behavior for c++ versions less than c++17: The left hand side of the = operator can be executed before or after the right hand side. In c++17, it's guaranteed that the right side will be executed before the left side. I think you assumed that in X[tot-1] tot's value was increased by 1 because of X[++tot], but actually tot is increasing later.