Submission #446156

#	Time	Username	Problem	Language	Result	Execution time	Memory
446156		abitpal	Balloons (CEOI11_bal)	C++14	100 / 100	1911 ms	10244 KiB

bal

#include <bits/stdc++.h>
using namespace std; 
 
#define ll long long
const ll MOD = 1e9 + 7; 
 
int di[] = {1, 0, -1, 0}, dj[] = {0, 1, 0, -1}; 
 
unsigned ll gcd(unsigned ll a ,unsigned ll b){
    if(b==0) return a;
    a%=b;
    return gcd(b,a);
}
 
unsigned ll lcm(unsigned ll a ,unsigned ll b){
    return (a /(gcd(a, b))) * b; 
}
 
ll binpow(ll a, ll b, ll m){
    a %= m; 
    ll ans = 1; 
    while (b > 0){
        if (b%2){
            ans = ans * a % m; 
        }
        a = a * a % m; 
        b /= 2; 
    }
    return ans % m; 
}

int e[2000]; 

int get(int x) {return e[x] < 0 ? x : e[x] = get(e[x]); }
int size(int x) {return -e[get(x)]; }
int unite(int a, int b){
    if (get(a) == get(b)) return 0; 
    a = get(a); b = get(b); 
    if (e[a] > e[b]) swap(a, b); 
    e[a] += e[b]; e[b] = a; 
    return 1; 
}


// dp[i] - number of ways to produce the sum of i, using the values on a dice
// dp[i] += dp[i-1] + dp[i-2] .. + dp[i-6]

double calc(pair<double, double> a, double b){
    double dif = pow(abs(a.first - b), 2); 
    return (dif/(4 * a.second)); 
}

int main(){

    int n; cin >> n; 
    pair<double,double> ar[n]; 

    for (int i = 0; i < n; i++){
        cin >> ar[i].first >> ar[i].second; 
    }

    double ans[n]; 
    ans[0] = ar[0].second; 

    vector<pair<double, double>> v; 
    v.push_back({ar[0].first, ar[0].second}); 

    for (int i = 1; i < n; i++){
        double cur = ar[i].second; 
        double x = ar[i].first; 
        for (auto i : v){
            cur = min(cur, calc(i, x)); 
        }

        if (cur != ar[i].second)
        while (!v.empty() && calc(v.back(), x) != cur) v.pop_back();
        while (!v.empty() && v.back().second <= cur) v.pop_back();
        v.push_back({x, cur});
        ans[i] = cur;  
    }

    for (int i = 0; i < n; i++){
        cout << fixed;
        cout << setprecision(3);
        cout << ans[i] << endl; 
    }

}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10