A

定义$a_{n+1}=a_n+maxdigit*mindigit$
尝试推导更多项后发现会出现$mindigit=0,a_{n+1}=a_n$的情况

int main()
{ios::sync_with_stdio(false); cin.tie(0);int t; cin >> t;while (t--) {ll a, k; cin >> a >> k; k--;while (k) {ll mind = a % 10, maxd = a % 10, b = a;while (b) {mind = min(mind, b % 10);maxd = max(maxd, b % 10);b /= 10;}k--;if (mind * maxd == 0) break;a += mind * maxd;}cout << a << endl;}
}

B

定义$a_i$表示第$i$个人需要一对的最少人数
贪心：排序后从小到大，满足即组一队

int a[maxn];
int main()
{ios::sync_with_stdio(false); cin.tie(0);int _; cin >> _;while (_--) {int n; cin >> n;for (int i = 0; i < n; ++i) {cin >> a[i];}sort(a, a + n);int mx = 0, now = 0, ans = 0;for (int i = 0; i < n; ++i) {now++;mx = max(mx - 1, a[i] - now);if (mx == 0) {now = 0;ans++;}			}cout << ans << endl;}
}

C

求满足$A\le x\le B\le y\le C\le z\le D$的三角形$(x,y,z)$数量
枚举x的长度，求y的范围和z的范围相乘

$x+y>z\rightarrow x+y\ge z+1\rightarrow z\le x+y-1$

• 当$x+y-1\le D$时，此时y的范围为$max(B,C+1-x)\le y\le min(C, D+1-x)$，z的范围为$x+y-C$，则答案为$\sum\limits_{y=l}^rx+y-C=(x - C) * (r - l + 1) + (l + r) * (r - l + 1) / 2;$
• 当$x+y-1>D\rightarrow x+y-2\ge D$，此时y的范围为$max(B,D+2-x)\le y\le C$，z的范围为$D-C+1$，则答案为$\sum\limits_{y=l}^CD-C+1$

typedef long long ll;
int main()
{ios::sync_with_stdio(false); cin.tie(0);ll A, B, C, D; while (cin >> A >> B >> C >> D) {ll sum = 0;for (int x = A; x <= B; ++x) {if (D + 1 - x >= B) {ll r = min(C, D + 1 - x);ll l = max(B, C + 1 - x);sum += (x - C) * (r - l + 1) + (l + r) * (r - l + 1) / 2;}if (D + 2 - x <= C) {ll t = max(B, D + 2 - x);sum += (D - C + 1) * (C - t + 1);}}cout << sum << endl;}
}

D

n个数的和为S，求设置一个k，没有子序列的和为k或者S-k
设置$n-1$个1，1个$S-n+1$，因此子序列不能构成的数范围为$(n,S-n)$，判断$n\le S-n$即可

int main()
{ios::sync_with_stdio(0); cin.tie(0);int n, s;while (cin >> n >> s) {int ans = s - n;if (ans >= n) {cout << "YES" << '\n';for (int i = 0; i < n - 1; ++i) {cout << 1 << ' ';}cout << ans + 1 << ' ' << n << endl;}else cout << "NO" << endl;}
}

E

给出n个围墙的高度$h_i$，可以花费$A$元使$h_i+1$，花费$R$元使$h_i-1$，花费$M$元使$h_i-1,h_j+1$。其中操作3等效于操作1和操作2.
发现根据确定的高度$x$，花费$f(x)$成一元二次函数，使用三分确定极值

int h[maxn];
ll n, a, r, m;
ll check(int x) {ll less = 0, great = 0;for (int i = 0; i < n; ++i) {if (h[i] < x) less += x - h[i];if (h[i] > x) great += h[i] - x;}if (m <= a + r) {ll minval = min(less, great);ll sum = minval * m;less -= minval; great -= minval;sum += less * a + great * r;return sum;}else {ll sum = less * a + great * r;return sum;}return 0;
}
int main()
{ios::sync_with_stdio(false); cin.tie(0);while (cin >> n >> a >> r >> m) {for (int i = 0; i < n; ++i) cin >> h[i];int l = 0, r = 1e9, ans;while (l <= r) {int lmid = l + (r - l) / 3;int rmid = r - (r - l) / 3;if (check(lmid) < check(rmid)) ans = lmid, r = rmid - 1;else ans = rmid, l = lmid + 1;}cout << check(ans) << endl;}
}