Daiwa Securities Co. Ltd. Programming Contest 2023（AtCoder Beginner Contest 331）

A - Tomorrow

int main() {
    IOS;
    for (_ = 1; _; --_) {
        int M, D, y, m, d; cin >> M >> D >> y >> m >> d;
        ++d;
        if (d > D) ++m, d -= D;
        if (m > M) ++y, m -= M;
        cout << y << ' ' << m << ' ' << d;
    }
    return 0;
}

B - Buy One Carton of Milk

int main() {
    IOS;
    for (_ = 1; _; --_) {
        int n, m, s, l; cin >> n >> s >> m >> l;
        memset(d, 0x3f, sizeof d); d[0] = 0;
        rep (i, 1, n + 11) {
            if (i - 6 >= 0) umin(d[i], d[i - 6] + s);
            if (i - 8 >= 0) umin(d[i], d[i - 8] + m);
            if (i - 12 >= 0) umin(d[i], d[i - 12] + l);
        }
        int mi = 1e9;
        rep (i, n, n + 11) umin(mi, d[i]);
        cout << mi;
    }
    return 0;
}

C - Sum of Numbers Greater Than Me

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n;
        rep (i, 1, n) cin >> a[i], b[i] = c[i] = i;
        sort(b + 1, b + 1 + n, [&](int x, int y) { return a[x] < a[y]; });
        rep (i, 1, n) s[i] = s[i - 1] + a[b[i]];
        c[b[n]] = n;
        per (i, n - 1, 1) {
            if (a[b[i]] == a[b[i + 1]]) c[b[i]] = c[b[i + 1]];
            else c[b[i]] = i;
        }
        rep (i, 1, n) cout << s[n] - s[c[i]] << ' ';
    }
    return 0;
}

D - Tile Pattern

没有修改操作，脑抽用二维树状算面积和了, 就不放板子了，用二维前缀和维护面积和就好了。

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> k;
        int t = n, s = 0; n <<= 1; m = n;
        rep (i, 1, t) rep (j, 1, t) {
            char c; cin >> c;
            if (c ^ 'W') {
                add(i, j, i, j, 1);
                add(i + t, j, i + t, j, 1);
                add(i, j + t, i, j + t, 1);
                add(i + t, j + t, i + t, j + t, 1);
                ++s;
            }
        }
        rep (i, 1, k) {
            int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
            int xx1 = x1 % t, yy1 = y1 % t, xx2 = x2 % t, yy2 = y2 % t;
            if (xx2 < xx1) xx2 += t;
            if (yy2 < yy1) yy2 += t;
            int x = (x2 - x1 - (xx2 - xx1)) / t, y = (y2 - y1 - (yy2 - yy1)) / t;
            ll res = ask(xx1 + 1, yy1 + 1, xx2 + 1, yy2 + 1);
            res += 1ll * x * y * s;
            res += ask(xx1 + 1, 1, xx2 + 1, t) * y + ask(1, yy1 + 1, t, yy2 + 1) * x;
            cout << res << '\n';
        }
    }
    return 0;
}

E - Set Meal

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m >> k;
        rep (i, 1, n) cin >> a[i], c[i] = i;
        rep (i, 1, m) cin >> b[i], d[i] = i;
        sort(c + 1, c + 1 + n, [&](int x, int y) { return a[x] > a[y]; });
        sort(d + 1, d + 1 + m, [&](int x, int y) { return b[x] > b[y]; });
        unordered_map<int, unordered_set<int>> st;
        rep (i, 1, k) {
            int x, y; cin >> x >> y;
            st[x].insert(y);
        }
        int mx = -1;
        for (int i = 1, mj = n + 1, j = 1; i <= n && j < mj; ++i, j = 1) {
            for (; j < mj; ++j) if (!st[c[i]].count(d[j])) {
                mj = j;
                umax(mx, a[c[i]] + b[d[j]]);
                break;
            }
        }
        cout << mx;
    }
    return 0;
}

F - Palindrome Query

线段树，哈希维护回文判断

const int N = 1e6 + 5, P = 13331;

int n, m, _, k, cas;
unsigned int p[N];
char s[N];

struct BitTree {
    struct node { 
        int l, r, len;
        unsigned int val1, val2;
    } tr[N << 2];
    void push_up(int rt) {
        tr[rt].val1 = tr[rt << 1].val1 * p[tr[rt << 1 | 1].len] + tr[rt << 1 | 1].val1;
        tr[rt].val2 = tr[rt << 1 | 1].val2 * p[tr[rt << 1].len] + tr[rt << 1].val2;
    }
    void build(int rt, int l, int r, char* a) {
        tr[rt].l = l, tr[rt].r = r, tr[rt].len = r - l + 1, tr[rt].val1 = tr[rt].val2 = 0;
        if (l == r) { tr[rt].val1 = tr[rt].val2 = a[l]; return; }
        int mid = l + r >> 1;
        build(rt << 1, l, mid, a); build(rt << 1 | 1, mid + 1, r, a);
        push_up(rt);
    }
    void change(int rt, int k, char c) {
        if (tr[rt].l == k && tr[rt].r == k) {
            tr[rt].val1 = tr[rt].val2 = c;
            return;
        }
        int mid = tr[rt].l + tr[rt].r >> 1;
        if (k <= mid) change(rt << 1, k, c);
        else if (k > mid) change(rt << 1 | 1, k, c);
        push_up(rt);
    }
    unsigned int ask(int rt, int l, int r, bool f) {
        if (r < l) return 0;
        if (tr[rt].l >= l && tr[rt].r <= r) return f ? tr[rt].val2 : tr[rt].val1;
        int mid = tr[rt].l + tr[rt].r >> 1;
        unsigned int ansl = l <= mid ? ask(rt << 1, l, r, f) : 0;
        unsigned int ansr = r > mid ? ask(rt << 1 | 1, l, r, f) : 0;
        unsigned int ans = 0;
        if (!f) ans = r > mid ? ansl * p[min(r, tr[rt << 1 | 1].r) - mid] + ansr : ansl;
        else ans = l <= mid ? ansr * p[mid - max(l, tr[rt << 1].l) + 1] + ansl : ansr;
        return ans;
    }
} bit;

int main() {
    IOS;
    for (_ = 1; _; --_) {
        cin >> n >> m >> s + 1; p[0] = 1;
        rep (i, 1, n) p[i] = p[i - 1] * P;
        bit.build(1, 1, n, s);
        rep (i, 1, m) {
            int op; cin >> op;
            if (op == 2) {
                int l, r; cin >> l >> r;
                int len = r - l + 1 >> 1;
                unsigned int x = bit.ask(1, l, l + len - 1, 0), y = bit.ask(1, r - len + 1, r, 1);
                cout << (x == y ? "Yes\n" : "No\n");
            } else {
                int k; char c; cin >> k >> c;
                bit.change(1, k, c);
            }
        }
    }
    return 0;
}