Submission #973849

#	Time	Username	Problem	Language	Result	Execution time	Memory
973849		GrindMachine	Fish (IOI08_fish)	C++17	0 / 100	503 ms	65536 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;

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> mnl(m+5,inf2);
    rep1(i,n) amin(mnl[a[i].ss],a[i].ff);

    vector<bool> came(m+5);
    bool bad[m+5][m+5];
    memset(bad,0,sizeof bad);
    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]--;
            ptr--;
        }

        if(came[x]) conts;
        came[x] = 1;
        
        ll prod = 1;
        rep1(y,m){
            if(bad[x][y]) conts;
            if(y != x and mnl[y] <= len/2){
                bad[y][x] = 1;
            }

            if(y != x){
                if(mnl[y] <= len/2){
                    prod = prod*(cnt[y]+1)%MOD;
                }
            }
            else{
                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: 0 / 4
• Subtask 2: 0 / 4
• Subtask 3: 0 / 4
• Subtask 4: 0 / 4
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5
• Subtask 7: 0 / 6
• Subtask 8: 0 / 6
• Subtask 9: 0 / 6
• Subtask 10: 0 / 6
• Subtask 11: 0 / 7
• Subtask 12: 0 / 7
• Subtask 13: 0 / 7
• Subtask 14: 0 / 5
• Subtask 15: 0 / 5
• Subtask 16: 0 / 5
• Subtask 17: 0 / 5
• Subtask 18: 0 / 5
• Subtask 19: 0 / 5