Is it precision error or something other?

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I know that this problem supposed to be a trivial problem , but I stay in it about 1 hour and i don't know why it get WA on test 3 about 10 times of submissions and i submitted the solution with python and c++ as i thought that WA is due to precision error but both of them didn't get accepted.
Any help please?

c++ code

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
//using namespace __gnu_pbds;
#define pb push_back
#define pf push_front
#define all(v) (v).begin() , (v).end()
#define allr(v) (v).rbegin() , (v).rend()
#define fi first
#define se second
#define forn(i,x,n) for(ll i=x;i<n;i++)
#define forr(i,n,x) for(ll i=n;i>x;i--)
#define fast ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define ones(n) __builtin_popcountll(n)
#define sum(a) accumulate(a.begin(),a.end(),0ll)
#define str(a) to_string(a)
#define len(a) (ll)a.size()

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
//template<class T> using order_set= tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

int main()
{
    fast;
    int n;
    cin>>n;
    forn(_,0,n)
    {
        ld temp,dist;
        cin>>temp>>dist;
        cout<<fixed<<setprecision(2)<<(dist * 1000.0) / (331.3 + .606 * temp)<<'\n';

    }
}

python code

for _ in range(int(input())):
    temp, dist = map(float, input().split())
    print(f'{(dist * 1000) / (331.3 + .606 * temp):.2f}')

Комментарии (6)

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Charizard2021

4 года назад, скрыть # |

It is about rounding, you have to set value = (dist * 1000.0) / (331.3 + .606 * temp), and then to round you have to floor(value*100+0.5)/100; Then it gets AC.

→ Ответить

beethoven97

4 года назад, скрыть # ^ |

thank you, my code get accepted in python and c++ also.

lotusblume

This still gives me wrong answer. What am I doing wrong?

Code

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main() {
    int tt;
    cin >> tt;
    while(tt--) {
        long double temp, dist;
        cin >> temp >> dist;
        long double value = (dist * 1000.0) / (331.3 + 0.606 * temp);
        value = floor(value * 100 + 0.5) / 100;
        cout << fixed << setprecision(2) << value << '\n';
    }
    return 0;
}

You don't use setprecision after rounding, due to it changing the answer. It gets AC when you remove fixed and setprecision(2).

Thank you :)

your code will be accepted if you print value without setprecision and fixed, but i don't know why these happened.

accepted code

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main() {
    int tt;
    cin >> tt;
    while(tt--) {
        long double temp, dist;
        cin >> temp >> dist;
        long double value = (dist * 1000.0) / (331.3 + 0.606 * temp);
        value = floor(value * 100 + 0.5) / 100;
        cout<<value<<'\n';
    }
    return 0;
}

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