GSS3 Description 动态维护最大子段和,支持单点修改. Solution 设 \(f[i]\) 表示以 \(i\) 为结尾的最大子段和, \(g[i]\) 表示 \(1 \sim i\) 的最大子段和,那么 \[f[i] = max(f[i - 1] + a[i], a[i])\] \[g[i] = max(g[i - 1], f[i])\] 发现只跟前一项有关.我们希望使用矩阵乘法的思路,但是矩阵乘法通常只能适用于递推问题.因此我们引入广义矩阵乘法. 矩阵乘法问题可分治的原因在于…

SP1716 GSS3 - Can you answer these queries III 题意翻译 n 个数,q 次操作 操作0 x y把A_xAx 修改为yy 操作1 l r询问区间[l, r] 的最大子段和 依旧是维护最大子段和,还是再敲一遍比较好. code: #include<iostream> #include<cstdio> #define ls(o) o<<1 #define rs(o) o<<1|1 using namespace std…

题意翻译 nnn 个数, qqq 次操作 操作0 x y把 AxA_xAx​ 修改为 yyy 操作1 l r询问区间 [l,r][l, r][l,r] 的最大子段和 题目描述 You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th ele…

问题描述 [LG-SP1716](https://www.luogu.org/problem/SP1716] 题解 GSS 系列的第三题,在第一题的基础上带单点修改. 第一题题解传送门 在第一题的基础上,增加一个单点修改就完事了. \(\mathrm{Code}\) #include<bits/stdc++.h> using namespace std; template <typename Tp> void read(Tp &x){ x=0;char ch=1;int f…

题面 题解 相信大家写过的传统做法像这样:(这段代码蒯自Karry5307的题解) struct SegmentTree{ ll l,r,prefix,suffix,sum,maxn; }; //... inline void update(ll node) { ll res; tree[node].sum=tree[node<<1].sum+tree[(node<<1)|1].sum; tree[node].maxn=max(tree[node<<1].maxn,tr…

SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for giv…

GSS3 - Can you answer these queries III You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for given x y print max{…

题意翻译 nnn 个数, qqq 次操作 操作0 x y把 AxA_xAx​ 修改为 yyy 操作1 l r询问区间 [l,r][l, r][l,r] 的最大子段和 感谢 @Edgration 提供的翻译 题目描述 You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:…

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it…

题目链接 之前用线段树写了一遍,现在用\(ddp\)再写一遍. #include <cstdio> #define lc (now << 1) #define rc (now << 1 | 1) inline int max(int a, int b){ return a > b ? a : b; } const int INF = 2147483647 >> 2; const int MAXN = 50010; inline int read(){…