제출 #737064

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
737064		GrindMachine	Nice sequence (IZhO18_sequence)	C++17	58 / 100	2070 ms	71248 KiB

sequence

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://codeforces.com/blog/entry/63991?#comment-478025
https://oj.uz/submission/682671


p(i) = pref sum until i
p(i) - p(i-n) < 0 => p(i) < p(i-n)
p(i) - p(i-m) > 0 => p(i) > p(i-m)

find a valid p and the answer can be restored from this

how to find a valid p
the only condition on p is the inequalities that we have

create an array that satisfies given inequalities:
in such cases, think toposort

add edge u-->v if p(u) < p(v)

add edges: i-->i-n and i-m-->i

find toposort, and assign p(i) values in the order every index appears in the toposort

how to find max len seq?
=> if we can make a seq of length k, then we can also make a seq of len k-1

so we can b.s on max len

*/

const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

vector<int> adj[N];
vector<bool> vis(N), recstack(N);
vector<int> topo;
bool cyc;

void dfs(int u) {
    vis[u] = 1;
    recstack[u] = 1;

    trav(v, adj[u]) {
        if (vis[v]) {
            if (recstack[v]) cyc = true;
            conts;
        }

        dfs(v);
    }

    topo.pb(u);
    recstack[u] = 0;
}

void solve(int test_case)
{
    int n, m; cin >> n >> m;

    auto ok = [&](int mid) {
        rep(i, mid + 1) adj[i].clear(), vis[i] = 0;

        rep1(i, mid) {
            if (i - n >= 0) {
                adj[i].pb(i - n);
            }

            if (i - m >= 0) {
                adj[i - m].pb(i);
            }
        }

        topo.clear();
        cyc = false;

        rep(i, mid + 1) {
            if (vis[i]) conts;
            dfs(i);
        }

        reverse(all(topo));

        if (cyc) return false;
        return true;
    };

    int l = 1, r = N - 1;
    int ans = 0;

    while (l <= r) {
        int mid = (l + r) >> 1;
        if (ok(mid)) {
            ans = mid;
            l = mid + 1;
        }
        else {
            r = mid - 1;
        }
    }

    ok(ans);

    vector<int> p(ans + 5);
    rep(i, sz(topo)) {
        p[topo[i]] = i;
    }

    cout << ans << endl;

    rep1(i, ans) {
        int val = p[i] - p[i - 1];
        cout << val << " ";
    }

    cout << endl;
}

int main()
{
    fastio;

    int t = 1;
    cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'void solve(int)':
sequence.cpp:30:36: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   30 | #define rep(i, n) for(int i = 0; i < n; ++i)
      |                                    ^
sequence.cpp:158:5: note: in expansion of macro 'rep'
  158 |     rep(i, sz(topo)) {
      |     ^~~

• Subtask 1: 6 / 6
• Subtask 2: 9 / 9
• Subtask 3: 14 / 14
• Subtask 4: 15 / 15
• Subtask 5: 14 / 14
• Subtask 6: 0 / 18
• Subtask 7: 0 / 24