Submission #973956

#	Time	Username	Problem	Language	Result	Execution time	Memory
973956		GrindMachine	Fish (IOI08_fish)	C++17	100 / 100	419 ms	44120 KiB

fish

#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) (int)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

/*

knew beforehand that this problem could be solved with segtree

*/

int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct segtree {
    // https://codeforces.com/blog/entry/18051

    /*=======================================================*/

    struct data {
        ll a;
    };

    data neutral = {1};

    data merge(data &left, data &right) {
        data curr;
        curr.a = left.a*right.a%MOD;
        return curr;
    }

    void create(int i, T v) {

    }

    void modify(int i, T v) {
        tr[i].a = v%MOD;
    }

    /*=======================================================*/

    int n;
    vector<data> tr;

    segtree() {

    }

    segtree(int siz) {
        init(siz);
    }

    void init(int siz) {
        n = siz;
        tr.assign(2 * n, neutral);
    }

    void build(vector<T> &a, int siz) {
        rep(i, siz) create(i + n, a[i]);
        rev(i, n - 1, 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    void pupd(int i, T v) {
        modify(i + n, v);
        for (i = (i + n) >> 1; i; i >>= 1) tr[i] = merge(tr[i << 1], tr[i << 1 | 1]);
    }

    data query(int l, int r) {
        data resl = neutral, resr = neutral;

        for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
            if (l & 1) resl = merge(resl, tr[l++]);
            if (!(r & 1)) resr = merge(tr[r--], resr);
        }

        return merge(resl, resr);
    }
};

void solve(int test_case)
{
    ll n,m; cin >> n >> m >> MOD;
    vector<pll> a(n+5);
    rep1(i,n) cin >> a[i].ff >> a[i].ss;
    sort(a.begin()+1,a.begin()+n+1);

    vector<ll> cnt(m+5);
    rep1(i,n) cnt[a[i].ss]++;

    vector<ll> ind(m+5,-1);
    vector<ll> ord;
    ord.pb(0);

    rev(i,n,1){
        ll x = a[i].ss;
        if(ind[x] != -1) conts;
        ind[x] = sz(ord);
        ord.pb(x);
    }

    segtree<ll> st(m+5);
    rep1(i,m) st.pupd(ind[i],cnt[i]+1);

    vector<bool> came(m+5);
    vector<ll> recent_del(m+5,-1);
    vector<ll> reach(n+5,-1);
    ll ptr = n;
    ll ans = 0;

    rev(i,n,1){
        auto [len,x] = a[i];
        while(ptr >= 1 and a[ptr].ff > len/2){
            cnt[a[ptr].ss]--;
            st.pupd(ind[a[ptr].ss],cnt[a[ptr].ss]+1);
            recent_del[a[ptr].ss] = ptr;
            ptr--;
        }

        if(came[x]) conts;
        came[x] = 1;
        reach[x] = ptr;

        // not maxed out
        ll prod = 1;
        ll aft_prod = st.query(ind[x]+1,m).a;
        prod = cnt[x]*aft_prod%MOD;

        // rep1(y,m){
        //     if(y != x and came[y]) conts;

        //     if(y != x){
        //         prod = prod*(cnt[y]+1)%MOD;
        //     }
        //     else{
        //         prod = prod*cnt[y]%MOD;
        //     }
        // }

        ans = (ans+prod)%MOD;

        // maxed out
        prod = 1;
        ll l = 1, r = ind[x]-1;
        ll first_good = ind[x];

        while(l <= r){
            ll mid = (l+r) >> 1;
            ll y = ord[mid];
            if(reach[y] < recent_del[x]){
                first_good = mid;
                r = mid-1;
            }
            else{
                l = mid+1;
            }
        }

        prod = st.query(first_good,ind[x]-1).a*aft_prod%MOD; 

        // rep1(y,m){
        //     if(x == y) conts;
        //     if(came[y] and reach[y] >= recent_del[x]) conts;            
        //     prod = prod*(cnt[y]+1)%MOD;
        // }

        ans = (ans+prod)%MOD;
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

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

    return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 4 / 4
• Subtask 4: 4 / 4
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• Subtask 14: 5 / 5
• Subtask 15: 5 / 5
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• Subtask 18: 5 / 5
• Subtask 19: 5 / 5