Submission #138029

#	Time	Username	Problem	Language	Result	Execution time	Memory
138029		qkxwsm	Remittance (JOI19_remittance)	C++14	100 / 100	628 ms	9916 KiB

remittance

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define INF 1000000007
#define LLINF 2696969696969696969ll
#define MAXN 1000013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N;
int A[MAXN], B[MAXN];

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> N;
	FOR(i, 0, N)
	{
		cin >> A[i] >> B[i];
	}
	//ok i think we can get away with trying to send stuff in batches of big
	FORD(i, 30, 1)
	{
		FOR(k, 0, 3)
		{
			FOR(j, 0, N)
			{
				if (A[j] >= B[j] + (1 << i))
				{
					A[j] -= (1 << i);
					A[(j == N - 1 ? 0 : j + 1)] += (1 << (i - 1));
				}
			}
		}
		// cerr << "HAVE:";
		// FOR(j, 0, N)
		// {
		// 	cerr << ' ' << A[j];
		// }
		// cerr << endl;
	}
	FOR(k, 0, 3)
	{
		FOR(i, 0, N)
		{
			if (A[i] >= 2 && A[i] > B[i])
			{
				A[i] -= 2;
				A[(i == N - 1 ? 0 : i + 1)]++;
			}
		}
	}
	FOR(i, 0, N)
	{
		if (A[i] != B[i])
		{
			cout << "No\n";
			return 0;
		}
	}
	cout << "Yes\n";
	return 0;
	//basically you're just checking if these two things are identical among these rules
}

• Subtask 1: 15 / 15
• Subtask 2: 40 / 40
• Subtask 3: 45 / 45