제출 #645120

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
645120		BackNoob	Vudu (COCI15_vudu)	C++14	140 / 140	296 ms	33504 KiB

vudu

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define endl '\n'
#define MASK(i) (1LL << (i))
#define ull unsigned long long
#define ld long double
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define BIT(x , i) ((x >> (i)) & 1)
#define TASK "task"
#define sz(s) (int) (s).size()

using namespace std;
const int mxN = 1e6 + 227;
const int inf = 1e9 + 277;
const int mod = 1e9 + 7;
const ll infll = 1e18 + 7;
const int base = 307;
const int LOG = 20;

template <typename T1, typename T2> bool minimize(T1 &a, T2 b) {
    if (a > b) {a = b; return true;} return false;
}
template <typename T1, typename T2> bool maximize(T1 &a, T2 b) {
    if (a < b) {a = b; return true;} return false;
}

struct FenwickTree{
    int n;
    vector<int> bit;
    FenwickTree(){}
    FenwickTree(int _n)
    {
        n = _n;
        bit.resize(n + 7, 0);
    }
    void update(int pos, int val)
    {
        for(; pos <= n; pos += pos & -pos) bit[pos] += val;
    }
    int getsum(int pos)
    {
        int res = 0;
        for(; pos > 0; pos -= pos & -pos) res += bit[pos];
        return res;
    }
} ft;

int n;
int a[mxN];
int p;
ll pre[mxN];

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

    for(int i = 1; i <= n; i++) a[i] -= p;

    for(int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];

    vector<ll> val;
    for(int i = 0; i <= n; i++) val.pb(pre[i]);
    sort(all(val));
    val.erase(unique(all(val)), val.end());

    ft = FenwickTree(n + 1);
    for(int i = 0; i <= n; i++) pre[i] = lower_bound(all(val), pre[i]) - val.begin() + 1;

    ll res = 0;
    for(int i = 1; i <= n; i++) {
        ft.update(pre[i - 1], 1);
        res += ft.getsum(pre[i]);
    }

    cout << res;
}


int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    //freopen("graph.inp", "r", stdin);
    //freopen("graph.out", "w", stdout);

    int tc = 1;
    //cin >> tc;
    while(tc--) {
        solve();
    }
    return 0;
}


/*
[lef[i], i] la thoa
[i, rig[i]] la thoa
i -> rig[i]
*/

• Subtask 1: 140 / 140