Submission #236774

#	Time	Username	Problem	Language	Result	Execution time	Memory
236774		ne4eHbKa	Krov (COCI17_krov)	C++17	70 / 140	567 ms	4852 KiB

krov

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 998244353;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 1e5 + 5;

int n;
int a[N];

void input() {
    cin >> n;
    fo(n) cin >> a[i];
}

struct tree {
    vi v, c, fc; ve<ll> fv; int n; ll s;
    tree(): v(0), c(0), fc(0), fv(0), n(0), s(0) {};
    void init() {
        c = v;
        sort(bnd(c));
        fc.assign(sz(v) + 5, 0);
        fv.assign(sz(v) + 5, 0);
    }
    void fcupd(int i, int v) { for(i++; i < sz(fc); i += i & -i) fc[i] += v; }
    void fvupd(int i, int v) { for(i++; i < sz(fv); i += i & -i) fv[i] += v; }
    int fcget(int i) { int v {}; for(i++; i; i -= i & -i) v += fc[i]; re v; }
    int fvget(int i) {  ll v {}; for(i++; i; i -= i & -i) v += fv[i]; re v; }
    int index(int v) { re upper_bound(bnd(c), v) - c.begin() - 1; }
    void add(int i) {
        int j = index(v[i]);
        fcupd(j, +1);
        fvupd(j, +v[i]);
        ++n, s += v[i];
    }
    void del(int i) {
        int j = index(v[i]);
        fcupd(j, -1);
        fvupd(j, -v[i]);
        --n, s -= v[i];
    }
    ll get(int c) {
        int j = index(c);
        if(j < 0) re s - ll(n) * c;
        ll le = fcget(j);
        ll ri = n - le;
        ll sle = fvget(j);
        ll sri = s - sle;
        re le * c - sle +
           sri - ri * c;
    }
};

void solve() {
    input();

    ll ans = +OO;

    tree le {}, ri {};

    fo(n) le.v.pb(a[i] - i); le.init();
    fo(n) ri.v.pb(a[i] + i); ri.init();

    for(int i = 0; i < n; ++i) ri.add(i);

    for(int i = 0; i < n; ++i) {
        ll l = max(i + 1, n - i), r = 2e9 + 5;
        while(l < r) {
            ll m0 = (l + r) >> 1;
            ll m1 = m0 + 1;
            ll f0 = le.get(m0 - i) + ri.get(m0 + i);
            ll f1 = le.get(m1 - i) + ri.get(m1 + i);
            if(f0 == f1) l = r = m0;
            if(f0 < f1) r = m0;
            if(f1 < f0) l = m1;
        }
        umin<ll>(ans, le.get(l - i) + ri.get(l + i));
        le.add(i);
        ri.del(i);
    }

    cout << ans << endl;
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
		cerr << "case #" << test << endl;
        solve();
        cerr << endl;
    }
#else
//    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

Compilation message (stderr)

krov.cpp: In function 'int m_bpow(ll, ll)':
krov.cpp:50:26: warning: suggest parentheses around '-' inside '<<' [-Wparentheses]
     for(ll p = 1ll << 63 - __builtin_clzll(b); p; p >>= 1) {
                       ~~~^~~~~~~~~~~~~~~~~~~~
krov.cpp: In function 'void solve()':
krov.cpp:9:18: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
 #define fr(i, n) for(int i = 0; i < n; ++i)
                  ^
krov.cpp:10:15: note: in expansion of macro 'fr'
 #define fo(n) fr(i, n)
               ^~
krov.cpp:118:5: note: in expansion of macro 'fo'
     fo(n) le.v.pb(a[i] - i); le.init();
     ^~
krov.cpp:118:30: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
     fo(n) le.v.pb(a[i] - i); le.init();
                              ^~
krov.cpp:9:18: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
 #define fr(i, n) for(int i = 0; i < n; ++i)
                  ^
krov.cpp:10:15: note: in expansion of macro 'fr'
 #define fo(n) fr(i, n)
               ^~
krov.cpp:119:5: note: in expansion of macro 'fo'
     fo(n) ri.v.pb(a[i] + i); ri.init();
     ^~
krov.cpp:119:30: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
     fo(n) ri.v.pb(a[i] + i); ri.init();
                              ^~

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 14 / 14
• Subtask 4: 0 / 14
• Subtask 5: 14 / 14
• Subtask 6: 0 / 14
• Subtask 7: 14 / 14
• Subtask 8: 0 / 14
• Subtask 9: 0 / 14
• Subtask 10: 0 / 14