Submission #941590

#	Time	Username	Problem	Language	Result	Execution time	Memory
941590		pan	Packing Biscuits (IOI20_biscuits)	C++17	100 / 100	8 ms	1372 KiB

biscuits

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include "bits_stdc++.h"
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
using namespace std;
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, srb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> piii;
typedef pair<pi, pi> pii;

long long count_tastiness(long long x, std::vector<long long> a)
{

	
	while (a.size()<=60) a.pb(0);
	//for (ll i=0; i<=60; ++i) show(a[i]);
	ll dp[65], ps[65];
	// prefix sum
	ps[0] = a[0];
	for (ll i=1; i<=60; ++i){ ps[i] = ps[i-1]/2 + a[i];}
	// base case
	if (ps[0]>=x) dp[0] = 2;
	else dp[0] = 1;
	// transition 
	for (ll i=1; i<=60; ++i)
	{
		dp[i] = dp[i-1]; // if ith bit turn off - always possible
		//show(dp[i]);
		if (ps[i]<x) continue;// not possible to turn ith on
		// now consider if we want to turn ith bit on
		ll on = 0;
		ll needed = max(0LL, x-a[i]); // cannot be used later, all neg become 0
		for (ll j=i-1; j>=1; --j)
		{
			needed*=2; // sig val halved
			if (ps[j]>=needed+x) // possible to turn this bit on
			{
				on+=dp[j-1]; // if we choose to turn off, freely choose any bit from 0 - i-1: done
				// now consider if we want to turn it on
				needed= max(0LL, needed + x- a[j]);
			}
			else // consider if can only turn it off
			{
				needed  = max(0LL, needed -a[j]);
			}
		}
		on+= (ps[0]>=needed*2+x?2:1); // base case
		dp[i] += on; // now, both 0, 1 cases considered
		//show(dp[i]);
		
	}
	return dp[60];
	
}

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