题意 从左到右选择四列，使得四列分别对应字母 $v,i,k,a$

string solve() {int n, m; cin >> n >> m;char s[N][N];char a[5] = "vika";int cnt = 0;for (int i = 0; i < n; i ++) cin >> s[i];for (int j = 0; j < m; j ++)for (int i = 0; i < n; i ++)if (s[i][j] == a[cnt]) {cnt ++;break;}if (cnt == 4) return yes;return no;
}

题意 从 $a$ 序列经过转换变成 $b$ 序列，已知 $b$ 序列，求任意一个可能的 $a$ 序列

首先 $b_0\leftarrow a_0$，然后只要满足 $a_{i-1}\le a_i$ 就将 $a_{i}push\_back$ 到 $b$ 序列后面

void solve() {int n; cin >> n;vector<int> a(1);cin >> a[0];for (int i = 1; i < n; i ++) {int x; cin >> x;if (*(a.end() - 1) > x) a.push_back(x);a.push_back(x);}cout << a.size() << endl;for (auto i : a) cout << i << ' ';cout << endl;
}

题意 给出每个栅栏的高度（宽度为 $1$），问这个栅栏是否关于 $y=x$ 对称

string solve() {int n; cin >> n;vector<int> a(n);for (auto &i : a) cin >> i;a.push_back(0);int idx = 0;for (int i = n - 1; i >= 0; i --) {int t = a[i] - a[i + 1];while (t --) if (a[idx ++] != i + 1) return no;}return yes;
}

题意 每两种球可以合成一种冰淇淋，且 $(i,j)=(j,i)$。特殊的，两个球 $1$ 可以合成 $(1,1)$。问至少需要多少球能刚好合成 $n$ 个冰淇淋

ll solve() {ll n; cin >> n;ll l = 0, r = 2e9;while (l < r) {ll mid = l + r + 1 >> 1;if (mid * (mid - 1) / 2 <= n) l = mid;else r = mid - 1;}ll res = l + (n - l * (l - 1) / 2);return res;
}

题意 一共 $n$ 天每天一场电影，最多看 $m$ 场电影，每场电影的热度为 $a_i$，每次看电影的兴奋值为 $a_i-cnt\cdot d$，其中 $d$ 为给定常数，$cnt$ 为距上次看电影的天数，求最终的兴奋值之和

题解 优先队列

ll solve() {int n, m, d; cin >> n >> m >> d;priority_queue<int, vector<int>, greater<int>> q;ll res = 0, sum = 0;for (int i = 1; i <= n; i ++) {int x; cin >> x;if (x <= 0) continue;if (q.size() < m) {sum += x;q.push(x);} else {sum += x;q.push(x);sum -= q.top();q.pop();}res = max(res, sum - (ll)i * d);}return res;
}

题意 有 $n$ 个怪兽血量为 $a_i$，魔法师初始有 $0$ 点水魔法和 $0$ 点火魔法，每秒分别恢复 $w,f$ 点，对于每个怪兽只能

法一：$bitset$

int solve() {int n, w, f; cin >> w >> f >> n;bitset<N> b;b[0] = true;int sum = 0;for (int i = 1; i <= n; i ++) {int x; cin >> x;sum += x;b |= b << x;}int res = sum;for (int i = 0; i <= sum; i ++) {if (!b[i]) continue;int x = (i + w - 1) / w;int y = (sum - i + f - 1) / f;res = min(res, max(x, y));}return res;
}

法二：$dp$ 因为是从二维优化到一维，因此 $j$ 逆序枚举

int solve() {int n, w, f; cin >> w >> f >> n;vector<int> a(n);int sum = 0;for (auto &i : a) cin >> i, sum += i;vector<int> dp(sum + 1);dp[0] = 1;for (auto i : a)for (int j = sum; j >= i; j --)dp[j] |= dp[j - i];int res = sum;for (int i = 0; i <= sum; i ++) {if (!dp[i]) continue;int x = (i + w - 1) / w;int y = (sum - i + f - 1) / f;res = min(res, max(x, y));}return res;
}