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Home/ Questions/Q 9127711
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T07:20:46+00:00 2026-06-17T07:20:46+00:00

I did something like this long double n; cin >> n; n = n

  • 0

I did something like this

long double n;

cin >> n;

n = n * 10000;

long long int temp = (long long) n;

now when i try to print temp, then a problem occours in some test cases like 2.36

for 2.36 the value of temp should be 23600 but the value of temp comes out to be 23599

Pls someone help me out with this already got 4 wrong ans for this.. small problem

for simplification ..
my code goes like this

int main()

{

int t;

for(scanf("%d", &t); t-- ;) {

    float n;
    scanf("%f", &n);
    n *= 10000;
    long int as = (long int) n;
   cout << "\nas : " << as << " n : " << n << endl;
    long  a, b;
    a = as;
    b = 10000;
    while(a%b != 0) {
        long  temp = a % b;
        a = b;
        b = temp;
    }
    long  factor = b;
    cout << (10000/factor) << endl;
}
return 0;

}

the aim of this program was something that .. i was given a number that can have at max 4 places after the decimal. that was the average score scored by a batsman so we had to find the minimum number of matches he should play to get that score

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  1. Editorial Team
    Editorial Team
    2026-06-17T07:20:48+00:00Added an answer on June 17, 2026 at 7:20 am

    This is because of the way floating-points are represented internally. You should round them off before performing truncation.

    Doing a floor(n+0.5) or a ceil(x-0.5) would round the number correctly.

    EDIT:

    As your truncation step is a floor(..) operation in itself, you should just do n = n * 10000 + 0.5 as @Mooing Duck stated.

    (Example)

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