公司让我做网站负责人,电子商务网站建设需要什么,合肥 网站运营,阿里云建站视频教程[ICPC2021 Nanjing R] Klee in Solitary Confinement 题面翻译 给定 n , k n,k n,k 和一个长为 n n n 的序列#xff0c;你可以选择对区间 [ l , r ] [l, r] [l,r] 的数整体加上 k k k#xff0c;也可以不加。最大化众数出现次数并输出。 题目描述 Since the travele…[ICPC2021 Nanjing R] Klee in Solitary Confinement 题面翻译 给定 n , k n,k n,k 和一个长为 n n n 的序列你可以选择对区间 [ l , r ] [l, r] [l,r] 的数整体加上 k k k也可以不加。最大化众数出现次数并输出。 题目描述 Since the traveler comes, People in Monstadt suddenly raise great interest in computer programming and algorithms, including Klee, the Spark Knight of the Knights of Favonius. Being sent to solitary confinement by Jean again, Klee decides to spend time learning the famous Mo’s algorithm, which can compute with a time complexity of O ( n 1.5 ) \mathcal{O}(n^{1.5}) O(n1.5) for some range query problem without modifications. To check whether Klee has truly mastered the algorithm (or in fact making another bombs secretly), Jean gives her a problem of an integer sequence a 1 , a 2 , ⋯ , a n a_1, a_2, \cdots, a_n a1​,a2​,⋯,an​ along with some queries [ l i , r i ] [l_i, r_i] [li​,ri​] requiring her to find the mode number in the contiguous subsequence a l i , a l i 1 , ⋯ , a r i a_{l_i}, a_{l_i 1}, \cdots, a_{r_i} ali​​,ali​1​,⋯,ari​​. The mode number is the most common number (that is to say, the number which appears the maximum number of times) in the subsequence. With the help of Mo’s algorithm, Klee solves that problem without effort, but another problem comes into her mind. Given an integer sequence a 1 , a 2 , ⋯ , a n a_1, a_2, \cdots, a_n a1​,a2​,⋯,an​ of length n n n and an integer k k k, you can perform the following operation at most once: Choose two integers l l l and r r r such that 1 ≤ l ≤ r ≤ n 1 \le l \le r \le n 1≤l≤r≤n and add k k k to every a i a_i ai​ where l ≤ i ≤ r l \le i \le r l≤i≤r. Note that it is OK not to perform this operation. Compute the maximum occurrence of the mode number of the whole sequence if you choose to perform (or not perform) the operation optimally. 输入格式 There is only one test case in each test file. The first line of the input contains two integers n n n and k k k ( 1 ≤ n ≤ 1 0 6 1 \le n \le 10^6 1≤n≤106, − 1 0 6 ≤ k ≤ 1 0 6 -10^6 \le k \le 10^6 −106≤k≤106) indicating the length of the sequence and the additive number. The second line of the input contains n n n integers a 1 , a 2 , ⋯ , a n a_1, a_2, \cdots, a_n a1​,a2​,⋯,an​ ( − 1 0 6 ≤ a i ≤ 1 0 6 -10^6 \le a_i \le 10^6 −106≤ai​≤106) indicating the original sequence. 输出格式 Output one line containing one integer indicating the maximum occurrence of the mode number of the whole sequence after performing (or not performing) the operation. 样例 #1 样例输入 #1 5 2 2 2 4 4 4样例输出 #1 5样例 #2 样例输入 #2 7 1 3 2 3 2 2 2 3样例输出 #2 6样例 #3 样例输入 #3 7 1 2 3 2 3 2 3 3样例输出 #3 5样例 #4 样例输入 #4 9 -100 -1 -2 1 2 -1 -2 1 -2 1样例输出 #4 3提示 For the first sample test case, choose l 1 l 1 l1 and r 2 r 2 r2 and we’ll result in the sequence { 4 , 4 , 4 , 4 , 4 } \{4, 4, 4, 4, 4\} {4,4,4,4,4}. The mode number is obviously 4 4 4 which appears 5 5 5 times. For the second sample test case, choose l 4 l 4 l4 and r 6 r 6 r6 and we’ll result in the sequence { 3 , 2 , 3 , 3 , 3 , 3 , 3 } \{3, 2, 3, 3, 3, 3, 3\} {3,2,3,3,3,3,3}. The mode number is 3 3 3 which appears 6 6 6 times. For the fourth sample test case, choose not to perform the operation. The mode number is 1 1 1 and − 2 -2 −2 which both appear 3 3 3 times. 以上来自洛谷 以上来自洛谷 以上来自洛谷 看完题目知道我为什么说”原神启动“了吧。什么不知道一看你就没看这篇题解。 重点声明我不玩原神。 解题思路 这一套 ICPC 的问题全是关于原神的诶出题人什么了 正片开始 我们枚举最后众数为 x x x则每次只需要单独考虑 x x x 和 x k xk xk。我们事先可以将每个数按数值大小将位置插入 vector则可做到均摊 O ( n ) O(n) O(n)。如果使用 m a p map map 或者别的容器实现则有运行超时的风险。 现在问题转化成有一个长度为 m m m 的序列序列仅由 X X X 和 Y Y Y 组成用 X l , r X_{l,r} Xl,r​ 和 Y l , r Y_{l,r} Yl,r​​ 表示区间 [ l , r ] [l,r] [l,r] 里 X X X 和 Y Y Y 的个数则我们需要选择一个区间 [ l , r ] [l,r] [l,r]使得 X 1 , l − 1 Y l , r X r 1 , m X_{1,l−1}Y_{l,r}X_{r1,m} X1,l−1​Yl,r​Xr1,m​ 最大。 简单转化一下则对于每一个 r r r我们需要最大化 X 1 , l − 1 Y l , r X r 1 , m X 1 , l − 1 ( r − l 1 ) − X l , r X r 1 , m X_{1,l−1}Y_{l,r}X_{r1,m}X_{1,l−1}(r−l1)−X_{l,r}X_{r1,m} X1,l−1​Yl,r​Xr1,m​X1,l−1​(r−l1)−Xl,r​Xr1,m​​。整理得到 ( 2 × X 1 , l − 1 − l ) ( r 1 − X 1 , r X r 1 , m ) (2\times X_{1,l−1}−l)(r1−X_{1,r}X_{r1,m}) (2×X1,l−1​−l)(r1−X1,r​Xr1,m​)即最大化 2 × X 1 , l − 1 − l 2\times X_{1,l−1}−l 2×X1,l−1​−l记录前缀最大值转移即可时间复杂度 O ( m ) O(m) O(m)。综上时间复杂度为 O ( n ) O(n) O(n)。 然后去写代码复制到提交代码出点击提交就会 A C AC AC ? AC Code #include bits/stdc.h using namespace std; #define int long long const int Maxn 1e6 5; int n, k, a[Maxn]; int tong[Maxn * 4][2], maxx, len; vectorint res[Maxn * 4]; int ans; inline void work() {cin n k;for (int i 1; i n; i) {cin a[i];a[i] 2e6;res[a[i]].push_back(a[i]);res[a[i] k].push_back(a[i]);len max(len, a[i] k);maxx max({maxx, (int)res[a[i]].size(), (int)res[a[i] k].size()});}if (!k) {cout maxx / 2 endl;return;}int tmp;for (int i 0; i len; i) {if (res[i].size() 0) {continue;}for (int j 0; j res[i].size(); j) {tong[j 1][0] tong[j][0] (res[i][j] i);tong[j 1][1] tong[j][1] (res[i][j] ! i);}tmp tong[1][0] - tong[1][1];for (int j 1; j res[i].size(); j) {tmp max(tmp, tong[j - 1][0] - tong[j - 1][1]);ans max(ans, tong[res[i].size()][0] tong[j][1] - tong[j][0] tmp);}}cout ans endl; } signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);work();return 0; }