Codeforces Round #707 (Div. 2, based on Moscow Open Olympiad in Informatics)

Codeforces Round #707 (Div. 2, based on Moscow Open Olympiad in Informatics)

A - Alexey and Train

阅读理解

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n; int cur = 0;
        rep (i ,1, n) cin >> a[i] >> b[i];
        rep (i, 1, n) {
            cin >> m;
            cur += (a[i] - b[i - 1] + m);
            if(i == n) break;
            cur += b[i] - a[i] + 1 >> 1;
            if (cur < b[i]) cur = b[i];
        }
        cout << cur << '
';
    }
    return 0;
}

B - Napoleon Cake

int main() {
    IOS;
    for (cin >> _; _; --_) {
        cin >> n;
        rep (i, 1, n) pre[i] = 0;
        rep (i, 1, n) {
            cin >> m; ++pre[max(1, i - m + 1)];
            --pre[i + 1];
        }
        rep (i, 1, n) pre[i] += pre[i - 1], cout << !!pre[i] << ' ';
        cout << '
';
    }
    return 0;
}

C - Going Home

假设找不到答案意味着, 最长的a[i] 为

1, 2(1 + 1), 4(2 + 2), 7(4 + 3), 11(7 + 5), 17(11 + 6), ....最长也就2237, 即当n > 2237 必定有解, 暴力判断就行

int n, m, _, k, cas;
int a[N];
unordered_map<int, PII> st;
 
int main() {
    IOS; cin >> n; rep (i ,1, n) cin >> a[i];
    rep (i, 1, n) rep (j, i + 1, n) {
        auto it = st.find(a[i] + a[j]);
        if (it == st.end()) st[a[i] + a[j]] = {i , j};
        else if(it->se.fi != i && it->se.fi != j && it->se.se != i && it->se.se != j)
            return cout << "YES
" << i << " " << j << " " << it->se.fi << " " << it->se.se << '
', 0;
    }
    cout << "NO
";
    return 0;
}

D - Two chandeliers

设 n < m, 预处理当c[i] 表示从a串的第i个位置开始匹配b串, 匹配到b串最后有多少个位置相同

在暴力找 n, m 的公倍数, 同时计算一圈公倍数能出多少错误, 直接取膜 k, 最后剩下的k, 在一圈公倍数内必定有解

在暴力跑一圈, 故复杂度为 O(n) 常数很大

const int N = 5e5 + 5;
 
ll n, m, _, k, cas;
int a[N], b[N], h[N << 1];
ll c[N];
 
int solve(int n, int m, int* a, int* b, int x, int k) {
    rep(i, 0, m - 1) {
        if (a[(x + i) % n] != b[i]) --k;
        if (!k) return i + 1;
    }
    return 0;
}
 
void work(int n, int m, int* a, int* b) {
    memset(h, -1, sizeof h);
    rep(i, 0, n - 1) h[a[i]] = i; ll d = 0, lcm = 0;
    rep(i, 0, m - 1) if (h[b[i]] != -1) ++c[((h[b[i]] - i) % n + n) % n];
    for (int be = 0; ;) {
        d += m - c[be]; be = (be + m % n) % n; lcm += m;
        if (!be) break;
    }
    ll ans = (k - 1) / d * lcm; k = k % d ? k % d : d;
    int be = 0;
    rep(i, 0, n - 1) {
        if (k > m - c[be]) k -= m - c[be], be = (be + m % n) % n, ans += m;
        else { cout << ans + solve(n, m, a, b, be, k) << '
'; return; }
    }
}
 
int main() {
    IOS; cin >> n >> m >> k;
    rep(i, 0, n - 1) cin >> a[i];
    rep(i, 0, m - 1) cin >> b[i];
    if (n <= m) work(n, m, a, b);
    else work(m, n, b, a);
    return 0;
}