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Can Anyone explain what is the precision error in Java

By ChatGPT-4o-mini, history, 3 months ago, In English

While solving problem B in Codeforces Round 976 when I was calculating my answer I was using the square root function to calculate

WA

but it was giving WA on test case on k = 854258779689358055

instead when I wrote the same code in C++ and submitted it , it gave AC....Could anyone tellme why that happened...

283627551

Help please...

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3 months ago, # |
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Change first line of the while loop:-

mid=L+(R-L)/2

You are experiencing the error because for worst case R=5e18 and L=5e18-1 So L+R=1e19-1 which exceeds the max value of long(9e18) and hence gives a faulty value(possibly -9e18 or something).

Hope this helped you (-:

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    3 months ago, # ^ |
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    It is still giving WA

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      3 months ago, # ^ |
      ← Rev. 4   Vote: I like it 0 Vote: I do not like it

      You should always use sqrtl instead of sqrt because of precision issues.

      Here is the equivalent

      static long sqrtl(long x) {

      long sq = (long) Math.sqrt(x);  // Take the integer part of the square root

    if (sq * sq > x) return sq - 1; // Ensure we don't overshoot the correct integer

    return sq;
}

      Hope this helps

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