Submission #206009

#	Time	Username	Problem	Language	Result	Execution time	Memory
206009		mode149256	Lamps (JOI19_lamps)	C++14	4 / 100	97 ms	57164 KiB

lamp

/*input
6
011011
110010

17
10110110110110101
01011100100110000

6
011011
110010
// 4
*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}
#pragma GCC optimize("Ofast")

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	str A, B;
	cin >> N >> A >> B;

	vii dp(N, vi(3, MOD));
	// 0 - done
	// 1 - ones
	// 2 - zeroes

	dp[0][0] = (A[0] != B[0]);
	dp[0][1] = 1 + (B[0] == '0');
	dp[0][2] = 1 + (B[0] == '1');

	for (int i = 1; i < N; ++i)
	{
		// pildom [0]
		dp[i][0] = min(dp[i][0], dp[i - 1][0] + (A[i] != B[i] and A[i - 1] == B[i - 1]));
		dp[i][0] = min(dp[i][0], dp[i - 1][1] + (A[i] != B[i] and B[i - 1] == '1'));
		dp[i][0] = min(dp[i][0], dp[i - 1][2] + (A[i] != B[i] and B[i - 1] == '0'));
		// dp[i][0] = min(dp[i][0], dp[i - 1][0] + (A[i] != B[i] and A[i - 1] == B[i - 1]));
		// dp[i][0] = min(dp[i][0], dp[i - 1][2] + (B[i] == '1' and B[i - 1] != '1'));

		// pildom [1]
		dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1 + (B[i] == '0'));
		dp[i][1] = min(dp[i][1], dp[i - 1][1] + (B[i] == '0' and B[i - 1] != '0'));
		// dp[i][1] = min(dp[i][1], dp[i - 1][2] + (B[i] == '0'));


		dp[i][2] = min(dp[i][2], dp[i - 1][0] + 1 + (B[i] == '1'));
		dp[i][2] = min(dp[i][2], dp[i - 1][2] + (B[i] == '1' and B[i - 1] != '1'));
	}

	// print(dp);
	// for (int i = 0; i < N; ++i) printf("%d", dp[i][0]); printf("\n");
	// for (int i = 0; i < N; ++i) printf("%d", dp[i][1]); printf("\n");
	// for (int i = 0; i < N; ++i) printf("%d", dp[i][2]); printf("\n");

	printf("%d\n", min(dp[N - 1][0], min(dp[N - 1][1], dp[N - 1][2])));
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 0 / 6
• Subtask 2: 0 / 41
• Subtask 3: 4 / 4
• Subtask 4: 0 / 49