水盐平衡

• 思维
• 判断一下浓度大小，再选择加水还是加盐。

signed main() {int T = 1;T = read();while (T--) {vector<int> a(5);for (int i = 1; i <= 4; ++i) a[i] = read();int t1 = a[1] * a[4], t2 = a[2] * a[3];t1 > t2? puts("S"): puts("Y");}return 0;
}

水平考试

• 思维
• 无论多选还是单选，有错的选项 $0$ 分，没选全满分（小灰灰会填满）。

signed main() {int T = 1;T = read();while (T--) {string s1, s2; cin >> s1 >> s2;bool f = 1;for (auto it: s1) {if (s2.find(it) == -1) {f = 0;break;}}if (s2.size() > 1) f? puts("10"): puts("0");else f? puts("10"): puts("0");}return 0;
}

数组段数

• 前缀和
• $pre{f}_i$ 表示从 $1$ 到 $i$ 最少划分多少段。
• 注意：如果 $a_i==a_{l-1}$ 你得保留从 $a_l$ 开始的那段。

signed main() {int T = 1;
//    T = read();while (T--) {int n = read(), m = read();vector<int> a(n + 1), pref(n + 1);for (int i = 1; i <= n; ++i) a[i] = read();for (int i = 1; i <= n; ++i) {if (a[i] ^ a[i - 1]) pref[i] = 1;pref[i] += pref[i - 1];}while (m--) {int l = read(), r = read();writeln(pref[r] - pref[l - 1] + (a[l - 1] == a[l]));}}return 0;
}

剪纸游戏

• $dfs$
• 题目说明了剪下来的图案不会联通，那么跑 $dfs$ 的时候就记录一下面积，同时筛选出最小的行，列和最大的行，列。如果是一个矩形那么 $S=(lin{e}_{max}-lin{e}_{min})\cdot (colum{n}_{max}-colum{n}_{min})$ ，判断并记录答案即可。

string mp[SN];
int n, m;
int d[4][2] = {0, -1, 0, 1, -1, 0, 1, 0};
bool vis[SN][SN];
tuple<int, int, int, int, int> dfs(int x, int y) {int x1 = x, y1 = y, x2 = x, y2 = y, sum1 = 1;for (int i = 0; i < 4; ++i) {int xx = x + d[i][0], yy = y + d[i][1];if (xx < 0 || yy < 0 || xx >= n || yy >= m || vis[xx][yy] || mp[xx][yy] == '*') continue;vis[xx][yy] = 1;auto [xs1, ys1, xs2, ys2, sum] = dfs(xx, yy);x1 = min(x1, xs1), y1 = min(y1, ys1);x2 = max(x2, xs2), y2 = max(y2, ys2);sum1 += sum;}return {x1, y1, x2, y2, sum1};
}
signed main() {int T = 1;
//    T = read();while (T--) {n = read(), m = read();for (int i = 0; i < n; ++i) cin >> mp[i];int ans = 0;for (int i = 0; i < n; ++i) {for (int j = 0; j < mp[i].size(); ++j) {if (mp[i][j] ^ '*' && !vis[i][j]) {vis[i][j] = 1;auto [x1, y1, x2, y2, sum] = dfs(i, j);if ((x2 - x1 + 1) * (y2 - y1 + 1) == sum) ++ans;}}}write(ans);}return 0;
}

可口蛋糕

• 前缀和 + 枚举 + 双指针 $O(n)$ $or$ 线段树 + 二分 $O(nlogn)$ 。
• 从左往右枚举右端点，在保证 $W$ 的情况下，你可以得出区间 $[1,j]$ ，在这个区间选择一个左端点然后减去 $1$ 到左端点的前缀和，因为是减去所以左端点的选定是区间 $[1,j]$ 的最小值的索引。
• 双指针左端点就保证 $W$ 的情况一直往右移动选出最小值就好，
• 注意：更新答案一定是满足 $W$ 的情况下才能更新。

线段树 $code$

vector<int> w(N), d(N), prefd(N), prefw(N);
struct p {int l, r, Min;
}tr[N << 2];
void pushup(int i) { tr[i].Min = min (tr[ls].Min, tr[rs].Min); }
void build(int i, int l, int r) {tr[i] = {l, r, (int)1e18};if (l == r) {tr[i].Min = prefd[l];return ;}int mid = (l + r) >> 1;build(ls, l, mid), build(rs, mid + 1, r);pushup(i);
}
int query(int i, int l, int r) {if (tr[i].l >= l && tr[i].r <= r) return tr[i].Min;int res = 1e18;if (tr[ls].r >= l) res = min(res, query(ls, l, r));if (tr[rs].l <= r) res = min(res, query(rs, l, r));return res;
}
signed main() {int T = 1;
//    T = read();while (T--) {int n = read(), W = read();for (int i = 1; i <= n; ++i) {prefw[i] = w[i] = read();prefw[i] += prefw[i - 1];}for (int i = 1; i <= n; ++i) {prefd[i] = d[i] = read();prefd[i] += prefd[i - 1];}build(1, 1, n);int res = 0, ans = -1e18, ww = 0;for (int i = 1; i <= n; ++i) {ww += w[i], res += d[i];if (ww >= W) {int pos = upper_bound(prefw.begin() + 1, prefw.begin() + i + 1, ww - W) - prefw.begin();if (!--pos) continue;ans = max({ans, res - query(1, 1, pos), res});}}write(ans);}return 0;
}

双指针 $code$

signed main() {int T = 1;
//    T = read();while (T--) {vector<int> w(N), d(N), prefd(N), prefw(N);int n = read(), W = read();for (int i = 1; i <= n; ++i) {prefw[i] = w[i] = read();prefw[i] += prefw[i - 1];}for (int i = 1; i <= n; ++i) {prefd[i] = d[i] = read();prefd[i] += prefd[i - 1];}int res = 0, ans = -1e18, ww = 0, l = 0, Min = 1e18;for (int i = 1; i <= n; ++i) {ww += w[i], res += d[i];while (ww - W >= prefw[l]) Min = min(Min, prefd[l++]);ans = max({ans, res - Min});}write(ans);}return 0;
}

喜欢序列

• $set$ + 树状数组 $or$ 线段树
• 预处理断点，$set$ 存断点下标。
• 考虑加上 $w$ 后 $l,r$ 处的关联情况。
• • $l$ 与前一个断点相邻，$r$ 不为断点。
• • $l$ 与前一个断点相邻，$r$ 为断点。
• • $l$ 与前一个断点不相邻，$r$ 不为断点。
• • $l$ 与前一个断点不相邻，$r$ 为断点。
• 在断点处的情况考虑更新后两个区间是否能合并即可。
• 不在断点处更新后则从端点处断开即可。
• 注意：断点下标可能为 $0orn$ 此时不用断开。

树状数组 $code$

vector<int> a(N), tr(N);
signed main() {auto add = [] (int x, int n, int v) {for (int i = x; i <= n; i += lowbit(i)) tr[i] += v;};auto query = [&] (int x, int n) {int ans = 0;for (int i = x; i; i -= lowbit(i)) ans += tr[i];return ans;};int T = 1;
//    T = read();while (T--) {int n = read(), m = read();for (int i = 1; i <= n; ++i) {a[i] = read();add(i, n, a[i] - a[i - 1]);}set<int> endpos;for (int i = 1; i <= n; ++i) {if (a[i] ^ (a[i - 1] + 1)) endpos.insert(i - 1);}endpos.insert(0), endpos.insert(n);int s = 0, pre = 0;for (auto it: endpos) {s += (it - pre) * (it - pre);pre = it;}auto powt = [&] (int x) {return x * x;};while (m--) {int l = read(), r = read(), w = read();if (!w) {writeln(s);continue;}add(l, n, w), add(r + 1, n, -w);auto pos1 = endpos.lower_bound(l), pos2 = endpos.lower_bound(r);if (l == *prev(pos1) + 1 && r ^ *pos2) {if (*prev(pos1)) {int v1 = query(l - 1, n), v2 = query(l, n);if (v1 + 1 == v2) {int t1 = powt(*pos1 - *prev(pos1)), t2 = powt(*prev(pos1) - *prev(prev(pos1)));int t3 = powt(*pos1 - *prev(prev(pos1)));s = s - t1 - t2 + t3;endpos.erase(l - 1);}}int t1 = powt(*pos2 - *prev(pos2));int t2 = powt(*pos2 - r), t3 = powt(r - *prev(pos2));s = s - t1 + t2 + t3;endpos.insert(r);}else if (l == *prev(pos1) + 1 && r == *pos2) {if (*prev(pos1)) {int v1 = query(l - 1, n), v2 = query(l, n);if (v1 + 1 == v2) {int t1 = powt(*pos1 - *prev(pos1)), t2 = powt(*prev(pos1) - *prev(prev(pos1)));int t3 = powt(*pos1 - *prev(prev(pos1)));s = s - t1 - t2 + t3;endpos.erase(l - 1);}}if (r ^ n) {int v1 = query(r, n), v2 = query(r + 1, n);if (v1 + 1 == v2) {int t1 = powt(*next(pos2) - r), t2 = powt(r - *prev(pos2));int t3 = powt(*next(pos2) - *prev(pos2));s = s - t1 - t2 + t3;endpos.erase(r);}}}else if (l ^ (*prev(pos1) + 1) && r == *pos2) {if (r ^ n) {int v1 = query(r, n), v2 = query(r + 1, n);if (v1 + 1 == v2) {int t1 = powt(*next(pos2) - r), t2 = powt(r - *prev(pos2));int t3 = powt(*next(pos2) - *prev(pos2));s = s - t1 - t2 + t3;endpos.erase(r);}}int t1 = powt(*pos1 - *prev(pos1));int t2 = powt(*pos1 - l + 1), t3 = powt(l - 1 - *prev(pos1));s = s - t1 + t2 + t3;endpos.insert(l - 1);}else if (l ^ (*prev(pos1) + 1) && r ^ *pos2) {int t1 = powt(*pos1 - *prev(pos1));int t2 = powt(*pos1 - l + 1), t3 = powt(l - 1 - *prev(pos1));s = s - t1 + t2 + t3;endpos.insert(l - 1);t1 = powt(*pos2 - *prev(pos2));t2 = powt(*pos2 - r), t3 = powt(r - *prev(pos2));s = s - t1 + t2 + t3;endpos.insert(r);}writeln(s);}}return 0;
}

线段树 $code$

vector<int> a(N);
struct p {int l, r, v, tag;
}tr[N << 2];
void build(int i, int l, int r) {tr[i] = {l, r, 0, 0};if (l == r) {tr[i].v = a[l];return ;}int mid = (l + r) >> 1;build(ls, l, mid), build(rs, mid + 1, r);
}
void pushdown(int i) {if (tr[i].tag) {tr[ls].tag += tr[i].tag, tr[rs].tag += tr[i].tag;tr[ls].v += (tr[ls].r - tr[ls].l + 1) * tr[i].tag;tr[rs].v += (tr[rs].r - tr[rs].l + 1) * tr[i].tag;tr[i].tag = 0;}
}
void modify(int i, int l, int r, int k) {if (tr[i].l >= l && tr[i].r <= r) {tr[i].tag += k;tr[i].v += k;return ;}pushdown(i);if (tr[ls].r >= l) modify(ls, l, r, k);if (tr[rs].l <= r) modify(rs, l, r, k);
}
int query(int i, int pos) {if (tr[i].l == tr[i].r) return tr[i].v;pushdown(i);if (pos <= tr[ls].r) return query(ls, pos);else return query(rs, pos);
}
signed main() {int T = 1;
//    T = read();while (T--) {int n = read(), m = read();for (int i = 1; i <= n; ++i) a[i] = read();build(1, 1, n);set<int> endpos;for (int i = 1; i <= n; ++i) {if (a[i] ^ (a[i - 1] + 1)) endpos.insert(i - 1);}endpos.insert(0), endpos.insert(n);int s = 0, pre = 0;for (auto it: endpos) {s += (it - pre) * (it - pre);pre = it;}auto powt = [&] (int x) {return x * x;};for (int i = 1; i <= m; ++i) {int l = read(), r = read(), w = read();if (!w) {writeln(s);continue;}modify(1, l, r, w);auto pos1 = endpos.lower_bound(l), pos2 = endpos.lower_bound(r);if (l == *prev(pos1) + 1 && r ^ *pos2) {if (*prev(pos1)) {int v1 = query(1, l - 1), v2 = query(1, l);if (v1 + 1 == v2) {int t1 = powt(*pos1 - *prev(pos1)), t2 = powt(*prev(pos1) - *prev(prev(pos1)));int t3 = powt(*pos1 - *prev(prev(pos1)));s = s - t1 - t2 + t3;endpos.erase(l - 1);}}int t1 = powt(*pos2 - *prev(pos2));int t2 = powt(*pos2 - r), t3 = powt(r - *prev(pos2));s = s - t1 + t2 + t3;endpos.insert(r);}else if (l == *prev(pos1) + 1 && r == *pos2) {if (*prev(pos1)) {int v1 = query(1, l - 1), v2 = query(1, l);if (v1 + 1 == v2) {int t1 = powt(*pos1 - *prev(pos1)), t2 = powt(*prev(pos1) - *prev(prev(pos1)));int t3 = powt(*pos1 - *prev(prev(pos1)));s = s - t1 - t2 + t3;endpos.erase(l - 1);}}if (r ^ n) {int v1 = query(1, r), v2 = query(1, r + 1);if (v1 + 1 == v2) {int t1 = powt(*next(pos2) - r), t2 = powt(r - *prev(pos2));int t3 = powt(*next(pos2) - *prev(pos2));s = s - t1 - t2 + t3;endpos.erase(r);}}}else if (l ^ (*prev(pos1) + 1) && r == *pos2) {if (r ^ n) {int v1 = query(1, r), v2 = query(1, r + 1);if (v1 + 1 == v2) {int t1 = powt(*next(pos2) - r), t2 = powt(r - *prev(pos2));int t3 = powt(*next(pos2) - *prev(pos2));s = s - t1 - t2 + t3;endpos.erase(r);}}int t1 = powt(*pos1 - *prev(pos1));int t2 = powt(*pos1 - l + 1), t3 = powt(l - 1 - *prev(pos1));s = s - t1 + t2 + t3;endpos.insert(l - 1);}else if (l ^ (*prev(pos1) + 1) && r ^ *pos2) {int t1 = powt(*pos1 - *prev(pos1));int t2 = powt(*pos1 - l + 1), t3 = powt(l - 1 - *prev(pos1));s = s - t1 + t2 + t3;endpos.insert(l - 1);t1 = powt(*pos2 - *prev(pos2));t2 = powt(*pos2 - r), t3 = powt(r - *prev(pos2));s = s - t1 + t2 + t3;endpos.insert(r);}writeln(s);}}return 0;
}