Submission #623978

#	Time	Username	Problem	Language	Result	Execution time	Memory
623978		jophyyjh	Packing Biscuits (IOI20_biscuits)	C++14	21 / 100	1096 ms	46228 KiB

biscuits

/**
 * Binary. Hmm, interesting. I started with [S2], which I solved with recursion in
 * O(kq). The basic idea is to treat two 2^i biscuits in the bag as one 2^(i+1).
 * 
 * A few greedy observations:
 * (1)  Given x, can we determine if it's possible? Well, we can simply iterate i
 *      from large to small. If the additional tastiness x we still need for a
 *      certain bag satisfies x >= 2^i, put a biscuit 2^i in the bag if there's one.
 * (2)  Biscuits of the same type can be distributed as evenly as possible, i.e.
 *      there is an optimal partition s.t. for every i
 *              |#2^i_biscuits_in_bag_1 - #2^i_biscuits_in_big_2| <= 1.
 * (3)  Unlike (1), (2), we consider from small to large. If x is odd, each bag needs
 *      at least 1 2^0 biscuit. After that, every 2 2^O biscuits can be grouped to
 *      form 1 2^1 biscuit, so we can add this num to the original num of 2^1
 *      biscuits. Well, let's write this recursively:
 *           def find():
 *               total = 0
 *               if #2^0 >= x:
 *                   add floor((#2^0 - x) / 2) to #2^1 and recurse downdwards
 *               add floor(#2^0 / 2) to 2^1 and recurse downwards
 *               ...
 * 
 * Can we AC this problem with (3)? I guess that the num of configurations given to
 * find() isn't too many, so we can optimize it with memoization. Indeed, we can
 * prove that in each level of find(), the num. of unique calls the next level of
 * find() is bounded by x. This is because x >= x/2 + x/4 + x/8 + x/16 + .... [S1-3]
 * are solved.
 * 
 * Time Complexity: O(xqk)      (partial solution)
 * Implementation 1             (Only solves [S1-3], may solve [S4])
*/

#include <bits/stdc++.h>
#include "biscuits.h"

typedef long long	ll;


ll x;
std::vector<ll> a;

struct pair_t {
    int p;
    ll first;
};

inline bool operator<(const pair_t& p1, const pair_t& p2) {
    return p1.p < p2.p || (p1.p == p2.p && p1.first < p2.first);
}

std::map<pair_t, ll> cache;

ll search(int p, ll first) {
    if (cache.find(pair_t{p, first}) == cache.end()) {
        int k = a.size();
        if (p == k - 1)
            return first / x + 1;
        ll total = search(p + 1, a[p + 1] + first / 2);
        if (first >= x)
            total += search(p + 1, a[p + 1] + (first - x) / 2);
        cache[pair_t{p, first}] = total;
    }
    return cache[pair_t{p, first}];
}

ll count_tastiness(ll _x, std::vector<ll> _a) {
    cache.clear();
    x = _x, a = _a;
    return search(0, a[0]);
}

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