//在gss1的基础上加了修改操作,一样的做法,加一个modify函数就可以了#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>usingnamespacestd;constintN=5e4+5;intn,m,type,l,r;intL,R,M,S;structTree {intl,r,mid;intsum,lmaxn,rmaxn,maxn; }tree[N<<2];intread...
第三行:一个正整数q,表示有q个询问; 第4~q+3行:每行三个数p,x,y,表示三种操作。 输出:对于每一个种类为1的询问,输出最大子段和。 输入样例: 4 1 2 3 4 4 1 1 3 0 3 -3 1 2 4 1 3 3 输出样例: 6 4 -3 解析:用线段树进行维护,记录下每一段的和(sum),最大子段和(maxv),最大前...
所以我们对线段树的每个节点引入一个标记cnt,用它来记录该节点对应的区域是否全部不需要开方,如果不需要开方,那么就直接return ,终止无效更新,从而提高效率。 1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5usingnamespacestd;67#defineLson rt<<1,l,mid8#defineRson rt<<1|1...
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We can answer the same question without all the complexity, by making role transfers a two-step process. A role transfer first designates a "pending" user to receive the role. The transfer is only completed after the "pending" user calls the corresponding "accept" method. ...
In case your query is invalid or you asked more thann2n2queries, program will print−1−1and will finish interaction. You will receiveWrong answerverdict. Make sure to exit immediately to avoid getting other verdicts. When you determine numbers in all cells, output "!". ...
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SP1716 GSS3 - Can you answer these queries III 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]....
0 上一篇: cf550D. Regular Bridge(构造) 下一篇: SPOJ2713GSS4 - Can you answer these queries IV(线段树) posted @ 2018-09-02 06:42 自为风月马前卒 阅读(454) 评论(0) 编辑 收藏 举报 登录后才能查看或发表评论,立即 登录 或者逛逛 博客园首页 【推荐】100%开源!大型工业跨平台软件C++源码提供...