AtCoder ABC189

这套题值得写一写，从C题开始就很有技巧

C - Mandarin Orange

给定一个数组 $a_1.....a_n$
对于每个 $a_i$ ，找到其左边第一个比他小的位置 $l_i,a_{l_i}<a_i$ ，找到其右边第一个比他小的位置 $r_i$ ，计算 $max((r_i-l_i -1)*a_i)$
找左边第一个比他小的位置可以用单调栈，手推一下就可以看出来
栈顶维护当前的数，把前面比它大的数都出栈，因为后面的数再去扫描时已经没有必要维护前面大的数了

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)def main():items = sys.version.split()fp = open("in.txt") if items[0] == "3.10.6" else sys.stdinn = int(fp.readline())a = list(map(int, fp.readline().split()))l = [0] * nr = [0] * nst = [-1]for i in range(n):while st[-1] >= 0 and a[st[-1]] >= a[i]:st.pop()l[i] = st[-1]st.append(i)st = [n]for i in range(n - 1, -1, -1):while st[-1] < n and a[st[-1]] >= a[i]:st.pop()r[i] = st[-1]st.append(i)ans = 0for i in range(n):t = (r[i] - l[i] - 1) * a[i]ans = max(t, ans)print(ans)if __name__ == "__main__":main()

D - Logical Expression

dp，记录当前操作后为0的种类和操作后为1的种类

# -*- coding: utf-8 -*-
# @time     : 2023/6/2 13:30
# @file     : atcoder.py
# @software : PyCharmimport bisect
import copy
import sys
from itertools import permutations
from sortedcontainers import SortedList
from collections import defaultdict, Counter, deque
from functools import lru_cache, cmp_to_key
import heapq
import math
sys.setrecursionlimit(100010)def main():items = sys.version.split()fp = open("in.txt") if items[0] == "3.10.6" else sys.stdinn = int(fp.readline())dp = [[0] * 2 for _ in range(n + 1)]dp[0][0] = dp[0][1] = 1for i in range(1, n + 1):s = fp.readline().strip()if s == "AND":dp[i][1] = dp[i - 1][1]dp[i][0] = dp[i - 1][1] + dp[i - 1][0] * 2else:dp[i][1] = 2 * dp[i - 1][1] + dp[i - 1][0]dp[i][0] = dp[i - 1][0]ans = dp[n][1]print(ans)if __name__ == "__main__":main()

E - Rotate and Flip

如果对坐标变换熟悉，那么很简单：
操作1： $(x, y) : (y, - x)$
$\left[ \begin{array}{ccc} 0 & -1\\ 1 & 0 \end{array} \right ] \times \left[ \begin{array}{ccc} x\\ y \end{array} \right ] = \left[ \begin{array}{ccc} y\\ -x \end{array} \right ]$
操作2：
$\left[ \begin{array}{ccc} 0 & 1\\ -1 & 0 \end{array} \right ] \times \left[ \begin{array}{ccc} x\\ y \end{array} \right ] = \left[ \begin{array}{ccc} -y\\ x \end{array} \right ]$
操作3：
$\left[ \begin{array}{ccc} -1 & 0\\ 0 & 1 \end{array} \right ] \times \left[ \begin{array}{ccc} x\\ y \end{array} \right ] + \left[ \begin{array}{ccc} 2p\\ 0 \end{array} \right ] = \left[ \begin{array}{ccc} 2p-x\\ y \end{array} \right ]$

操作4：
$\left[ \begin{array}{ccc} 1 & 0\\ 0 & -1 \end{array} \right ] \times \left[ \begin{array}{ccc} x\\ y \end{array} \right ] + \left[ \begin{array}{ccc} 0\\ 2p \end{array} \right ] = \left[ \begin{array}{ccc} x\\ 2p-y \end{array} \right ]$
每一步都左乘即可。需要维护的是
$A x + d$ 中两个矩阵 $A, d$

#define _CRT_SECURE_NO_WARNINGS#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;int n, m, q;
struct Query {int step, x, i;
};struct Mat {LL a[2][2] = {0};int r, c;
};Mat raw[200020];
vector<pii> trans;
Query query[200020];
pll ans[200020];bool cmp(Query& lhs, Query& rhs) {return lhs.step < rhs.step;
}Mat mul(Mat& a, Mat& b) {Mat ret;ret.r = a.r, ret.c = b.c;for (int i = 0; i < a.r; ++i) {for (int j = 0; j < b.c; ++j) {for (int k = 0; k < a.c; ++k) {ret.a[i][j] += a.a[i][k] * b.a[k][j];}}}return ret;
}int main() { //freopen("in.txt", "r", stdin);scanf("%d", &n);for (int i = 0; i < n; ++i) {LL x, y;scanf("%lld%lld", &x, &y);Mat mat;mat.r = 2, mat.c = 1;mat.a[0][0] = x, mat.a[1][0] = y;raw[i] = mat;}Mat m0;m0.r = m0.c = 2;m0.a[0][0] = 0, m0.a[0][1] = 1, m0.a[1][0] = -1, m0.a[1][1] = 0;Mat m1 = m0;m1.a[0][0] = 0, m1.a[0][1] = -1, m1.a[1][0] = 1, m1.a[1][1] = 0;Mat m2 = m0;m2.a[0][0] = -1, m2.a[0][1] = 0, m2.a[1][0] = 0, m2.a[1][1] = 1;Mat m3 = m0;m3.a[0][0] = 1, m3.a[0][1] = 0, m3.a[1][0] = 0, m3.a[1][1] = -1;scanf("%d", &m);for (int i = 0; i < m; ++i) {int op;scanf("%d", &op);op--;if (op == 0 || op == 1) {trans.push_back({ op, 0 });}else {int p;scanf("%d", &p);trans.push_back({ op, p });}}Mat cur;cur.r = cur.c = 2;cur.a[0][0] = 1, cur.a[1][1] = 1;Mat d0;d0.r = 2, d0.c = 1;scanf("%d", &q);for (int i = 0; i < q; ++i) {int p, x;scanf("%d%d", &p, &x);x--;query[i].step = p, query[i].x = x, query[i].i = i;}sort(query, query + q, cmp);int i = 0, j = 0;while (i <= m && j < q) {int step = query[j].step;int ri = query[j].i;int x = query[j].x;if (i == step) {Mat nm = mul(cur, raw[x]);ans[ri] = { nm.a[0][0] + d0.a[0][0], nm.a[1][0] + d0.a[1][0] };j += 1;}else {if (trans[i].first == 0) {cur = mul(m0, cur);d0 = mul(m0, d0);}else if (trans[i].first == 1) {cur = mul(m1, cur);d0 = mul(m1, d0);}else if (trans[i].first == 2) {LL p = trans[i].second;cur = mul(m2, cur);d0 = mul(m2, d0);d0.a[0][0] += p * 2;}else {LL p = trans[i].second;cur = mul(m3, cur);d0 = mul(m3, d0);d0.a[1][0] += p * 2;}i += 1;}}for (int i = 0; i < q; ++i) {printf("%lld %lld\n", ans[i].first, ans[i].second);}return 0;
}

F - Sugoroku2

本题的思路和强化学习中贝尔曼方程非常类似
当递推式中存在未知数，可以将未知数设为一个需要解的元，最后列方程求解。
具体到此题中，
$\\ f(i)=f(0), i \in A \\ f(i)=(f(i+1)+f(i+2)...f(i+m))/m +1, i\notin A$
由于只存在一个未知数，可以将其单独列出求系数
即
$f(i)=a_if(0)+b_i$
对两者进行dp递推
最后 $f(0)=a_0f(0)+b_0$
求解 $f (0)$

#define _CRT_SECURE_NO_WARNINGS#include <iostream>
#include <string>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#define LT(x) (x * 2)
#define RT(x) (x * 2 + 1)using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;int n, m, k;
double a[400040], b[400040], sa[400040], sb[400040];
set<int> ks;int main() { //freopen("in.txt", "r", stdin);scanf("%d%d%d", &n, &m, &k);for (int i = 0; i < k; ++i) {int t;scanf("%d", &t);ks.insert(t);}for (int i = n - 1; i >= 0; --i) {if (ks.count(i)) {a[i] = 1.0;b[i] = 0.0;}else {a[i] = (sa[i + 1] - sa[i + m + 1]) / m;b[i] = (sb[i + 1] - sb[i + m + 1]) / m + 1;}sa[i] = sa[i + 1] + a[i];sb[i] = sb[i + 1] + b[i];}if (1 - a[0] < 1e-7) {printf("-1\n");}else {printf("%.9f\n", b[0] / (1 - a[0]));}return 0;
}