Submission #891144

#	Time	Username	Problem	Language	Result	Execution time	Memory
891144		vjudge1	Sure Bet (CEOI17_sure)	C++17	100 / 100	76 ms	8528 KiB

sure

#include<bits/stdc++.h>

#pragma GCC target("avx2")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;

void dgb_out() { cerr << "\n"; }
template<typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) { cerr << " " << H; dgb_out(T...); }
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
 
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file(s) freopen(s".in", "r", stdin);freopen(s".out", "w", stdout);
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ft first
#define sd second
#define ll long long
#define pll pair<ll,ll>
const int N = 5e5 + 10;
const int M = 2e7 + 5;
const int B = 316;
const ll msize = 2;
const ll mod1 = 1e9 + 7;
const ll mod2 = 998244353;
const long double Phi = acos(-1);
const long long inf = 2e9;

ll binmul(ll x, ll ti, ll m);
ll binpow(ll x, ll ti, ll m);

ll n;
long double a[N], b[N];

const void solve() {
    cin >> n;
    for (int i = 1; i <= n; ++ i) cin >> a[i] >> b[i];

    sort(a + 1, a + n + 1); reverse(a + 1, a + n + 1);
    sort(b + 1, b + n + 1); reverse(b + 1, b + n + 1);

    int i = 1, j = 1;
    long double ans = 0, cnt = 0, suma = 0, sumb = 0;
    while (i <= n || j <= n) {
        if ((j <= n && suma >= sumb) || (i == n + 1)) {
            sumb += b[j ++];
        } else {
            suma += a[i ++];
        }
        ++ cnt;

        ans = max(ans, min(suma, sumb) - cnt);
    }
    cout << fixed << setprecision(4) << ans;
}  

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    srand(time(NULL));

    //file("monsters");

    int tt = 1;
    // cin >> tt;
    for (int i = 1; i <= tt; ++ i) {
        solve();
    }

    return 0;
} 


// Template functions
ll binmul(ll x, ll ti, ll m) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= m; res %= m;} return res;}
ll binpow(ll x, ll ti, ll m) { ll res = 1;while (ti){if(ti & 1)res=binmul(res,x,m);x=binmul(x,x,m);ti >>= 1; x %= m; res %= m;} return res;}

• Subtask 1: 20 / 20
• Subtask 2: 40 / 40
• Subtask 3: 40 / 40