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back to boardMy solution (RUS/ENG) (Accepted) #include <iostream> #include <iomanip> #include <cmath> using namespace std; int main() { int x1, y1, x2, y2, x0, y0, l; cin >> x1 >> y1 >> x2 >> y2; cin >> x0 >> y0 >> l; cout << fixed << setprecision(2); // <|ENG: Set 2 symbols after dot |> <|RUS: stavim 2 simvola dlia tochki |> //<|ENG: get all sides |> //<|RUS: poluchaem vse storonui |> double firstEdge = sqrt(pow(x1 - x0, 2) + pow(y1 - y0, 2)); double secondEdge = sqrt(pow(x2 - x0, 2) + pow(y2 - y0, 2)); double field = sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2)); double ans2 = max(firstEdge, secondEdge) - l; // <|ENG: Second answer - max length |> <|RUS: Vtoroi otvet = prosto maximum do krainih tochek |> double ans1; //<|ENG: get the squares of the sides|> //<|RUS: Vosvodim v kvadrat storonui |> double a = firstEdge * firstEdge; double b = secondEdge * secondEdge; double c = field * field; //<|ENG: checking if there are obtuse angles on the side that is the pineapple field |> //<|RUS: Proverka est li typie ygli vosle toi storoni kotoraiy iavliaetsia polem c ananasami |> if (a+c < b || b+c < a) { ans1 = min(firstEdge, secondEdge) - l; } else { //<|ENG: get half the perimeter of triangle|> //<|RUS: poluchaem poly perimetr treygolnika |> double p = (firstEdge + secondEdge + field) / 2; //<|ENG: get square of triangle|> //<|RUS: poluchaem ploshad treygolnika |> double S = sqrt(p * (p - firstEdge) * (p - secondEdge) * (p - field)); //<|ENG: If the area is zero and the field with pineapples does not exist (test: 2 2 2 2 3 3 1)|> //<|RUS: Esli ploshad ravna 0 i polia c ananasami kak bu net (test: 2 2 2 2 3 3 1)|> if (S == 0 && field == 0) { ans1 = min(firstEdge, secondEdge) - l; } //<|ENG: If the area is zero and more than one side is not zero (test: -5 0 5 0 1 0 1)|> //<|RUS: Esli ploshad ravna 0 i ni odna storona ne ravna 0 (test: -5 0 5 0 1 0 1)|> else if (S == 0 && firstEdge != 0 && secondEdge != 0 && field != 0) { ans1 = 0; } else { //<|ENG: Default. We just look for the height using the formula|> //<|RUS: Obichni slychai. Prosto po formule S = a*h/2 (test: -5 0 5 0 1 0 1)|> double d = 2 * S / field; ans1 = d - l; } } //<|ENG: If, after subtracting the rope, the length becomes negative, then do 0|> //<|RUS: Esli sle vichetania verevki otvet stal menihe nylia to delayem 0|> ans1 = ans1 < 0 ? 0 : ans1; ans2 = ans2 < 0 ? 0 : ans2; //<|ENG: Next just cout "round(ans1 * 100) / 100" and "round(ans2 * 100) / 100"|> //<|RUS: Teper prosto vivedi "round(ans1 * 100) / 100" i "round(ans2 * 100) / 100"|> //!!!!!!!!!!!!!!!!!!! //cout << round(ans1 * 100) / 100 << endl; //cout << round(ans2 * 100) / 100; return 0; //<|ENG: Thanks a lot for your time here!|> //<|RUS: Spasibo chto pochitali i uia vam pomog (navernoe)|> // :) } |
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