Submission #1154470

#	Time	Username	Problem	Language	Result	Execution time	Memory
1154470		pan	Lamps (JOI19_lamps)	C++20	100 / 100	67 ms	49456 KiB

lamp

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#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 all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//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>, rb_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> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll const INF = 1e15;
ll dp[1000005][3][2]; 

int main()
{
	ll n;
	string a,b;
	cin >> n >> a >> b;
	a = '?' + a, b = '?' + b;
	for (ll i=1; i<=n; ++i)
	{
		for (ll j=0; j<3; ++j) for (ll k=0; k<2; ++k) dp[i][j][k] = INF;
		for (ll j=0; j<3; ++j) for (ll k=0; k<2; ++k)
		{
			// j: 0 - no, 1 - '0', 2 - '1'
			// k: 0 - no, 2 - xor
			// Case: 
			// 0, 1 
			if (a[i]!=b[i]) dp[i][0][1] = min(dp[i][0][1], dp[i-1][j][k] + (j> 0));
			// set then toggle
			// 1, 1
			if (b[i]=='1') dp[i][1][1] = min(dp[i][1][1], dp[i-1][j][k] + (j==2));
			// 2, 1
			if (b[i]=='0') dp[i][2][1] = min(dp[i][2][1], dp[i-1][j][k] + (j==1));
			// no need to consider toggle then set as it can be modelled as set opposite then toggle
			// 0, 0
			if (a[i]==b[i]) dp[i][0][0] = min(dp[i][0][0], dp[i-1][j][k] + (j > 0) + (k > 0));
			// 1, 0
			if (b[i]=='1') dp[i][2][0] = min(dp[i][2][0], dp[i-1][j][k] + (j==1) + (k > 0));
			// 2, 0
			if (b[i]=='0') dp[i][1][0] = min(dp[i][1][0], dp[i-1][j][k] + (j==2) + (k > 0));
		}
	}
	ll ans = INF;
	for (ll j=0; j<3; ++j) for (ll k=0; k<2; ++k) ans = min(ans, dp[n][j][k] + (j > 0) + (k > 0));
	cout << ans << endl;
	return 0;
}

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