# Untitled

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```#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define nl "\n"
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define pb push_back
#define llmx LONG_LONG_MAX
#define llmn LONG_LONG_MIN
#define mod 1000000007
const int N = 1e5 + 7;
const int INF = 1e9 + 10;

bool predicate(double mid, ll x1, ll y1, ll x2, ll y2, ll ww, ll hh, ll w, ll h) {
y1 -= mid;
y2 -= mid;
ll rem_up = h - y2;
ll rem_down = y1 - 0;
if (rem_up >= hh || rem_down >= hh) return true;

y1 += mid * 2;
y2 += mid * 2;
ll rem_up2 = h - y2;
ll rem_down2 = y1 - 0;
if (rem_up2 >= hh || rem_down2 >= hh) return true;

y1 -= mid;
y2 -= mid;
x1 -= mid;
x2 -= mid;
ll rem_right = w - x2;
ll rem_left = x1 - 0;
if (rem_right >= ww || rem_left >= ww) return true;

x1 += mid * 2;
x2 += mid * 2;
ll rem_right2 = w - x2;
ll rem_left2 = x1 - 0;
if (rem_right2 >= ww || rem_left2 >= ww) return true;

return false;
}

void solve(){
ll w, h;
cin >> w >> h;
ll x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
ll ww, hh;
cin >> ww >> hh;

double l = 0;
double r = 1e8;
double ans = -1;
while (r - l > 1e-6) {
double mid = l + (r - l) / 2;
if (predicate(mid, x1, y1, x2, y2, ww, hh, w, h)) {
ans = mid;
r = mid - 1;
}
else {
l = mid + 1;
}
}
cout << ans << nl;
}

/* Hey you should check this out
* int overflow, array bounds
* reset global array and variable
* look for special cases (n=1?)
* do something instead of nothing and stay organized
* bruteforce to find pattern
* DON'T GET STUCK ON ONE APPROACH
* Think the problem backwards
* In practice time don't see failing test case
*/

signed main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while(t--) {
solve();
}
return 0;
}```