Submission #361792

#	Time	Username	Problem	Language	Result	Execution time	Memory
361792		ryansee	Packing Biscuits (IOI20_biscuits)	C++14	0 / 100	2 ms	364 KiB

biscuits

#include "biscuits.h"
 
#include "bits/stdc++.h"
using namespace std;
 
#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define f first
#define s second
#define cbr cerr<<"hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x,y) (lower_bound(all(x),y)-x.begin())
#define ubd(x,y) (upper_bound(all(x),y)-x.begin())
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) { return rng() % (y+1-x) + x; } //inclusive
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }
 
using ll=long long; 
using ld=long double;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
using pi=pair<ll,ll>; using spi=pair<ll,pi>; using dpi=pair<pi,pi>; 
 
long long LLINF = 1e18;
int INF = 1e9+1e6;
#define MAXN (200006)
long long count_tastiness(long long x, vector<long long> a) {
	ll n = siz(a);
	ll sum = 0;
	vector<ll> cnt(60, 0), ans(60, 0), store(60, 0);
	// vector<ll> v={0};
	// FOR(i,0,18ll) {
		// sum += (i < a.size() ? a[i] : 0) * (1ll<<i);
		// ll sz = siz(v);
		// FOR(j,0,sz-1) if(v[j] + (1ll<<i) <= sum / x) {
			// v.eb(v[j] + (1ll<<i));
		// }
	// }
	FOR(i,0,59) {
		if(i < n) sum += a[i] * (1ll<<i);
		ll minus = 0;
		if(i < n) minus = a[i] % x * (1ll<<i);
		ll left = (sum - minus)/ x - (1ll<<i);
		store[i] = left;
		if(left < 0) goto bruh;
		if(left >= (1ll<<i) - 1) {
			cnt[i] = i ? ans[i-1] : 1;
		} else {
			++ cnt[i];
			DEC(j,i-1,0) {
				if((1ll<<j) & left) {
					cnt[i] += j ? ans[j-1] : 1;
					left = min(left, store[j]);
					if(left < 0) goto bruh;
				}
			}
			++ cnt[i];
		}
		bruh:;
		ans[i] = (i ? ans[i-1] : 1) + cnt[i];
		// if(i <= 3) cerr<<sum<<' '<<left<<' '<<cnt[i]<<'\n';
	}
	return ans.back();
}

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