제출 #1052521

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1052521		ProtonDecay314	비스킷 담기 (IOI20_biscuits)	C++17	42 / 100	29 ms	10476 KiB

biscuits

#include "biscuits.h"
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
typedef vector<vpi> vvpi;
typedef vector<vpll> vvpll;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef short int si;
typedef vector<si> vsi;
typedef vector<vsi> vvsi;
#define IOS ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
#define L(varll, mn, mx) for(ll varll = (mn); varll < (mx); varll++)
#define LR(varll, mx, mn) for(ll varll = (mx); varll > (mn); varll--)
#define LI(vari, mn, mx) for(int vari = (mn); vari < (mx); vari++)
#define LIR(vari, mx, mn) for(int vari = (mx); vari > (mn); vari--)
#define INPV(varvec) for(auto& varveci : (varvec)) cin >> varveci
#define fi first
#define se second
#define pb push_back
#define INF(type) numeric_limits<type>::max()
#define NINF(type) numeric_limits<type>::min()
#define TCASES int t; cin >> t; while(t--)

ll count_tastiness(ll x, vll a) {
    vvll dp;

    ll MAX_X = 10'010ll;

    ll MAX_K = 60ll;

    for(ll i = 0ll; i <= MAX_K; i++) {
        vll dpr(MAX_X, 1ll);
        dp.pb(dpr);
    }

    // Extend a

    while((ll)a.size() < MAX_K) {
        a.pb(0ll);
    }

    // Remove excess

    for(ll i = 0ll; i < MAX_K; i++) {
        if(a[i] >= x + 2) {
            ll excess = a[i] - x;

            a[i + 1] += excess >> 1ll;

            a[i] -= (excess >> 1ll) << 1ll;
        }
    }

    for(ll i = MAX_K - 1ll; i >= 0; i--) {
        for(ll n = 0ll; n < MAX_X; n++) {
            ll aipn = a[i] + n;

            if(aipn >= x) {
                dp[i][n] = dp[i + 1][aipn >> 1ll] + dp[i + 1][(aipn - x) >> 1ll]; 
            } else {
                dp[i][n] = dp[i + 1][aipn >> 1ll];
            }
        } 
    }

    // cout << dp[0][0] << "\n";

    // cout << flush;

	return dp[0][0];
}

• Subtask 1: 9 / 9
• Subtask 2: 12 / 12
• Subtask 3: 21 / 21
• Subtask 4: 0 / 35
• Subtask 5: 0 / 23