제출 #1052533

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1052533		dozer	비스킷 담기 (IOI20_biscuits)	C++14	33 / 100	40 ms	83036 KiB

biscuits

#include "biscuits.h"
#include<bits/stdc++.h>
using namespace std;
#define fileio() freopen("input.txt", "r", stdin), freopen("output.txt", "w", stdout)
#define fastio() cin.tie(0), ios_base::sync_with_stdio(0)
#define sp " "
#define endl "\n"
#define pii pair<int, int>
#define st first
#define nd second
#define pb push_back
#define ll long long
#define LL node * 2
#define RR node * 2 + 1
#define N 130
#define M 20005

const int modulo = 1e9 + 7;
const ll INF = 2e18 + 7;
 
 
ll arr[N], X;
ll dp[N][M], done[N][M];

int n;

ll f(int pos, ll sum){
	if (pos >= N) return 1;
	if (done[pos][sum]) return dp[pos][sum];
 
	ll curr_sum = sum + arr[pos];
	ll t1 = 0, t2 = 0;
	if (curr_sum >= X) t1 = f(pos + 1, (curr_sum - X) / 2);
	t2 = f(pos + 1, curr_sum / 2);
    done[pos][sum] = 1;
    //cout<<pos<<sp<<sum<<sp<<t1<<sp<<t2<<endl;
	return dp[pos][sum] = t1 + t2;
}
 
 
long long count_tastiness(long long x, vector<long long> a) {
	n = a.size();
	X = x;
    for (int i = 0; i < (int)a.size(); i++){
        if (a[i] > x + 1){
            if (i < n - 1) a[i + 1] += (a[i] - x) / 2;
            else a.pb((a[i] - x) / 2);
            a[i] = a[i] - ((a[i] - x) / 2) * 2;
        }
    }  

    n = a.size();

    for (int i = 0; i < N; i++){
        arr[i] = 0;
        for (int j = 0; j <= 2 * x; j++)
            done[i][j] = 0;
    }
 
 

    for (int i = 0; i < n; i++)
        arr[i] = a[i];


	ll res = f(0, 0);
	return res;
}
 
/*
 
int main() {
	fileio();
	int q;
	assert(scanf("%d", &q) == 1);
	vector<int> k(q);
	vector<long long> x(q);
	vector<vector<long long>> a(q);
	vector<long long> results(q);
	for (int t = 0; t < q; t++) {
		assert(scanf("%d%lld", &k[t], &x[t]) == 2);
		a[t] = vector<long long>(k[t]);
		for (int i = 0; i < k[t]; i++) {
			assert(scanf("%lld", &a[t][i]) == 1);
		}
	}
	fclose(stdin);
 
	for (int t = 0; t < q; t++) {
		results[t] = count_tastiness(x[t], a[t]);
	}
	for (int t = 0; t < q; t++) {
		printf("%lld\n", results[t]);
	}
	fclose(stdout);
	return 0;
}*/

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