Submission #156997

#	Time	Username	Problem	Language	Result	Execution time	Memory
156997		atoiz	Lamps (JOI19_lamps)	C++14	4 / 100	30 ms	4528 KiB

lamp

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cassert>
#include <numeric>
#include <utility>
#include <tuple>
#include <bitset>
#include <climits>
#include <random>
#include <chrono>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <ios>
#include <iomanip>

using namespace std;

#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define FORA(i, a) for (auto &i : a)
#define FORB(i, a, b) for (int i = a; i >= b; --i)
#define SZ(a) ((int) a.size())
#define ALL(a) begin(a), end(a)

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

const int INF = 1e9;
int N;
string A, B;
int F[2][2][3];
void upd(int &a, int b) { if (a > b) a = b; }

signed main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> N >> A >> B;
    FOR(i, 0, 1) FOR(j, 0, 1) FOR(k, 0, 2) F[i][j][k] = INF;
    F[0][0][0] = 1;
    F[0][1][0] = (A[0] != B[0]);
    FOR(i, 0, N - 2) {
        int id = i & 1;
        // FOR(j, 0, 1) FOR(k, 0, 2) cerr << F[id][j][k] << ' '; cerr << endl;
        FOR(j, 0, 1) FOR(k, 0, 2) F[id ^ 1][j][k] = INF;

        if (B[i] == B[i + 1]) {
            FOR(k, 0, 2) upd(F[id ^ 1][0][k], F[id][0][k]);
        } else {
            upd(F[id ^ 1][0][1], F[id][0][0] + 1);
            upd(F[id ^ 1][0][2], F[id][0][1]);
            upd(F[id ^ 1][0][1], F[id][0][2] + 1);
        }
        FOR(k, 0, 2) upd(F[id ^ 1][0][0], F[id][1][k] + 1);

        if ((A[i] != B[i]) == (A[i + 1] != B[i + 1])) {
            FOR(k, 0, 2) upd(F[id ^ 1][1][k], F[id][1][k]);
        } else {
            upd(F[id ^ 1][1][1], F[id][1][0] + 1);
            upd(F[id ^ 1][1][2], F[id][1][1]);
            upd(F[id ^ 1][1][1], F[id][1][2] + 1);
        }
        FOR(k, 0, 2) upd(F[id ^ 1][1][0], F[id][0][k] + (A[i + 1] != B[i + 1]));
    }

    int ans = INF;
    FOR(j, 0, 1) FOR(k, 0, 2) upd(ans, F[(N - 1) & 1][j][k]);
    cout << ans << endl;
}

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