Wow! Such Sequence!(线段树4893)
Wow! Such Sequence!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3808 Accepted Submission(s): 1079
Problem Description
Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It’s a mysterious blackbox.
After some research, Doge found that the box is maintaining a sequence an of n numbers internally, initially all numbers are zero, and there are THREE “operations”:
1.Add d to the k-th number of the sequence.
2.Query the sum of ai where l ≤ i ≤ r.
3.Change ai to the nearest Fibonacci number, where l ≤ i ≤ r.
4.Play sound “Chee-rio!”, a bit useless.
Let F0 = 1,F1 = 1,Fibonacci number Fn is defined as Fn = Fn - 1 + Fn - 2 for n ≥ 2.
Nearest Fibonacci number of number x means the smallest Fn where |Fn - x| is also smallest.
Doge doesn’t believe the machine could respond each request in less than 10ms. Help Doge figure out the reason.
Input
Input contains several test cases, please process till EOF.
For each test case, there will be one line containing two integers n, m.
Next m lines, each line indicates a query:
1 k d - “add”
2 l r - “query sum”
3 l r - “change to nearest Fibonacci”
1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, |d| < 231, all queries will be valid.
Output
For each Type 2 (“query sum”) operation, output one line containing an integer represent the answer of this query.
Sample Input
1 1
2 1 1
5 4
1 1 7
1 3 17
3 2 4
2 1 5
Sample Output
0
22
Author
Fudan University
Source
2014 Multi-University Training Contest 3
线段树的好题
/*
本题所求的可以分为两部分,一是在原来的基础上进行加和,二是区间转化为最近的fib.前者线段树的单点更新,没有优化的程度,只能做单点更新,
后者是对一个区间的操作,不能进行的单纯的单点更新,否则会超时,所以可以利用lazy,将每一个点的fib算出来,区间的也算出来,当需要进行操
作3的时候,可以查询到所需要更改的区间将其fib的区间和赋给sum,但是这种操作会影响操作1,因为你只是更新的区间的值,而叶子节点的值是原
值,当你进行单点更新的时候,会在原值的基础上进行,所以得到结果是不对的开始的时候我用lazy标记区间的值是不是都相同,进行区间的合并,在
区间的更新的时候减少时间复杂度,节点记录的是相同的值,但是超时,而且代码还十分冗长,我想是在查询的时候超时了,后来用lazy标记这个区间
是不是进行操作3了,节点分别记录着区间和和fib区间的和,进行操作3的时候直接查询到所在的区间,将其sum=fibsum,并将这个区间的lazy进行
标记,但进行其他的操作的时候,只要判断所在区间lazy是否标记,如果标记就向下进行更新,同时将这个区间的lazy取消,表示这个操作在这个区间
已经完成,不会对子区间产生影响.在回溯的时候进行向上更新,维护每个每个节点意义的正确性
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <set>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MAX=100000+100;
struct node
{
LL num;//Fibsum
LL sum;//区间和
bool lazy;//标记区间是不是进行了操作3,
} Tree[MAX*5];
LL Arr[100];
void pushup(int site,int L,int R)//向上更新
{
if(L==R)//相等都时候没有必要处理
{
return ;
}
Tree[site].sum=Tree[site<<1].sum+Tree[site<<1|1].sum;
Tree[site].num=Tree[site<<1].num+Tree[site<<1|1].num;
}
void pushDown(int site,int L, int R)//向下更新
{
if(Tree[site].lazy)//如果区间标记
{
if(L==R)
{
Tree[site].sum=Tree[site].num;
}
else
{
Tree[site<<1].lazy=Tree[site<<1|1].lazy=true;
Tree[site<<1].sum=Tree[site<<1].num;
Tree[site<<1|1].sum=Tree[site<<1|1].num;
}
Tree[site].lazy=false;
}
}
LL Rearch(int L,int R,LL s)//二分查找上限
{
while(L<=R)
{
int mid=(L+R)>>1;
if(Arr[mid]>=s)
{
R=mid-1;
}
else
{
L=mid+1;
}
}
if(Arr[L]==s)
{
return Arr[L];
}
else
{
if(Arr[L]-s>=s-Arr[L-1])
{
return Arr[L-1];
}
else
{
return Arr[L];
}
}
}
void Build(int L,int R,int site)//建立线段树,进行初始化
{
if(L==R)
{
Tree[site].lazy=false;
Tree[site].num=1;
Tree[site].sum=0;
return ;
}
int mid=(L+R)>>1;
Build(L,mid,site<<1);
Build(mid+1,R,site<<1|1);
Tree[site].num=Tree[site<<1].num+Tree[site<<1|1].num;
Tree[site].sum=Tree[site<<1].sum+Tree[site<<1|1].sum;
Tree[site].lazy=false;
}
void update(int L,int R,int l,int r,int site)//进行操作3
{
int mid=(L+R)>>1;
if(L==l&&R==r)
{
Tree[site].lazy=true;
Tree[site].sum=Tree[site].num;
return ;
}
pushDown(site,L,R);
if(r<=mid)
{
update(L,mid,l,r,site<<1);
}
else if(l>mid)
{
update(mid+1,R,l,r,site<<1|1);
}
else
{
update(L,mid,l,mid,site<<1);
update(mid+1,R,mid+1,r,site<<1|1);
}
pushup(site,L,R);
}
void Add(int L,int R,int site,int k,int s)//进行操作2
{
if(L==R)
{
Tree[site].sum+=s;
Tree[site].num=Rearch(1,80,Tree[site].sum);
return ;
}
int mid=(L+R)>>1;
pushDown(site,L,R);
if(k<=mid)
{
Add(L,mid,site<<1,k,s);
}
else
{
Add(mid+1,R,site<<1|1,k,s);
}
pushup(site,L,R);
}
LL Query(int L,int R,int l,int r,int site)//进行操作2
{
if(L==l&&R==r)
{
return Tree[site].sum;
}
int mid=(L+R)>>1;
pushDown(site,L,R);
if(r<=mid)
{
return Query(L,mid,l,r,site<<1);
}
else if(l>mid)
{
return Query(mid+1,R,l,r,site<<1|1);
}
else
{
return Query(L,mid,l,mid,site<<1)+Query(mid+1,R,mid+1,r,site<<1|1);
}
}
int main()
{
int op,l,r,d;
Arr[0]=1;
Arr[1]=1;
for(int i=2;i<=80;i++)
{
Arr[i]=Arr[i-1]+Arr[i-2];
}
int n,m;
while(~scanf("%d %d",&n,&m))
{
Build(1,n,1);
for(int i=1;i<=m;i++)
{
scanf("%d",&op);
switch(op)
{
case 1:
scanf("%d %d",&l,&d);
Add(1,n,1,l,d);
break;
case 2:
scanf("%d %d",&l,&r);
printf("%I64d\n",Query(1,n,l,r,1));
break;
case 3:
scanf("%d %d",&l,&r);
update(1,n,l,r,1);
}
}
}
return 0;
}
Wow! Such Sequence!(线段树4893)的更多相关文章
- hdu 4893 Wow! Such Sequence!(线段树)
题目链接:hdu 4983 Wow! Such Sequence! 题目大意:就是三种操作 1 k d, 改动k的为值添加d 2 l r, 查询l到r的区间和 3 l r. 间l到r区间上的所以数变成 ...
- 2016暑假多校联合---Rikka with Sequence (线段树)
2016暑假多校联合---Rikka with Sequence (线段树) Problem Description As we know, Rikka is poor at math. Yuta i ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树 区间取摸
D. The Child and Sequence Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...
- hdu4893Wow! Such Sequence! (线段树)
Problem Description Recently, Doge got a funny birthday present from his new friend, Protein Tiger f ...
- 2014多校3 Wow! Such Sequence!段树
主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=4893 这个问题还真是纠结啊--好久不写线段树的题了.由于这几天学伸展树.然后认为线段树小case了. ...
- HDU 6047 Maximum Sequence(线段树)
题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=6047 题目: Maximum Sequence Time Limit: 4000/2000 MS (J ...
- Codeforces 438D The Child and Sequence - 线段树
At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at ...
- hdu 5828 Rikka with Sequence 线段树
Rikka with Sequence 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5828 Description As we know, Rik ...
- hdu-5805 NanoApe Loves Sequence(线段树+概率期望)
题目链接: NanoApe Loves Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 ...
随机推荐
- C/C++中的实参和形参
今天突然看到一道关于形参和实参的题,我居然不求甚解.藐视过去在我的脑海里只有一个参数的概念,对于形参和实参的区别还真的不知道,作为学习了几年C++的人来说,真的深深感觉对不起自己对不起C++老师 T ...
- LINUX数据库的备份,以及远程授权登陆
mysql dump -u root -p juhui > /data/juhui.sql //备份数据库 grant all privileges on *.* to xf111@loca ...
- 使用LVM对硬盘在线扩容
初始状态: root@control:/dev/nova-volumes# vgdisplay --- Volume group --- VG Name nova-volumes System ID ...
- 利用脚本获取mysql的tps,qps等状态信息
#!/bin/bash mysqladmin -uroot -p'123456' extended-status -i1|awk 'BEGIN{local_switch=0;print "Q ...
- JAVA程序改错 (易错题)
JAVA程序改错 1. abstract class Name { private String name; public abstract boolean isStupidName(String n ...
- Java学习-048-插件应用之 Find Bugs
FindBugs 是一个静态分析工具,它可以检查类或者 JAR 文件,将字节码与一组缺陷模式进行对比以发现可能的问题,使用 FindBugs 可以在不实际运行程序的情况对软件进行分析.使用时最好将字节 ...
- Struts2 框架下 session 读出来为null
我用的strust2框架,开始的时候这么写的: 在 登陆函数中(注释部分): public String dealerLogin(){ EntityInfo entityinfo=dea ...
- SSD果然劲爆!
前两周入手了一块浦科特128G盘,不说多了,有图为证 以前把机械盘放在主硬盘位的时候,鲁大师显示是SATA II接口,现在把SSD放在主硬盘位,显示居然是SATA III接口了,看上面测试,确实是II ...
- iOS/OS X线程安全的基础知识
处理多并发和可重入性问题,是每个库发展过程中面临的比较困难的挑战之一.在Parse平台上,我们尽最大的努力保证你在使用我的SDKs时所做的操作都是线程安全的,保证不会出现性能问题. 在这篇文章中我们将 ...
- shopnc导入商品到大商创
<?php //select member_name user_name,member_mobile mobile_phone,member_email email,member_passwd ...