牛客练习赛82

牛客练习赛82

A - Mocha 的字符串

int a[N], c, ans = 2e9;
char s[N], t[] = "mocha";

bool check(int x) {
    rep (i, 0, 4) if (s[i + x] ^ t[i]) return 0;
    return 1;
}

int main() {
    IOS; cin >> n >> k >> s + 1;
    rep (i, 1, n - 4) if (check(i)) {
        ++c; a[m] = i; m = i;
        if (c >= k) _ = a[_], umin(ans, i - _ + 5);
    }
    if (ans == 2e9) cout << "poor Mocha";
    else cout << "Mocha suki!
" << ans;
    return 0;
}

B - Mocha 的序列

真的服气, sb的想高精去了

仔细观察操作3, 把取膜操作改作除法, 像不像

(C_r^{r - l + 1}) 我们当然知道这是个整数

故取膜肯定为0

int a[N];

int main() {
    IOS; cin >> n >> m;
    rep (i, 1, n) a[i] = i;
    while (m--) {
        int op, l, r, k; cin >> op >> l >> r;
        if (op == 2) cin >> k;
        else if (op == 3) cout << "0
";
    }
    return 0;
}

C - Mocha 的二元关系

模拟即可,

先把(b_i == b_j, i eq j) 且 (c_i eq c_j) 的直接输出(-1)

然后就是挑出(c_i == c_j, i eq j) 找出位置(c_i) 能填的数, 也就是(b_i, b_j) 所属集合能填的数的并集

然后又要最小, 就能求出每个数最早出现的位置, 然后贪心填数即可

const int N = 1e5 + 5;

int n, m, _, k, cas;
int c[N], b[N];
VI lis[N], g[N];
set<int> h[N], cur;
bool v[N];

int main() {
    IOS; cin >> k >> n >> m;
    rep (i, 1, k) { int a, b; cin >> a >> b; lis[a].pb(b); }
    rep (i, 1, m) cin >> b[i];
    rep (i, 1, m) {
        cin >> _;
        if (c[b[i]] && c[b[i]] != _) return cout << -1, 0;
        c[b[i]] = _; b[i] = 0;
    }
    rep (i, 1, n) if (c[i]) {
        for (auto &j : lis[i]) {
            umax(b[j], c[i]);
            if (!v[c[i]]) h[c[i]].insert(j);
            else if (h[c[i]].count(j)) cur.insert(j);
        }
        if (v[c[i]]) h[c[i]] = cur, clear(cur);
        v[c[i]] = 1; 
    }
    rep (i, 1, n)
        if (b[i]) g[b[i]].pb(i);
        else cur.insert(i);
    rep (i, 1, n) {
        for (auto &j : g[i]) cur.insert(j);
        if (!v[i]) {
            if (cur.empty()) return cout << -1, 0;
            b[i] = *cur.begin(), cur.erase(cur.begin());
        } else {
            auto it = cur.end();
            for (auto &j : h[i]) if ((it = cur.find(j)) != cur.end()) break;
            if (it == cur.end()) return cout << -1, 0;
            b[i] = *it; cur.erase(it);
        }
    }
    rep (i, 1, n) cout << b[i] << ' ';
    return 0;
}