438D - The Child and Sequence
4 seconds
256 megabytes
standard input
standard output
At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.
Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:
- Print operation l, r. Picks should write down the value of
. - Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r).
- Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).
Can you help Picks to perform the whole sequence of operations?
The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space:a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.
Each of the next m lines begins with a number type
.
- If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
- If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2.
- If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3.
For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.
这题关键是证明啊。 好线段树。
详细证明见:http://codeforces.com/blog/entry/12513

;
typedef ],Max[MAX<<];
]+sum[o<<|];
Max[o]=max(Max[o<<],Max[o<<|]);
}
;
build(L,mid,o<<);
build(mid+,R,o<<|);
push_up(o);
}
;
,ls,rs,mod);
,R,o<<|,ls,rs,mod);
push_up(o);
}
;
,k,x);
,R,o<<|,k,x);
push_up(o);
}
LL Query(;
;
,ls,rs);
,R,o<<|,ls,rs);
) {
build(,n,);
;i<m;i++) {
scanf() {
scanf(,n,,l,r));
}
) {
scanf(,n,,l,r,x) ;
}
,n,,k,x);
}
}
}
;
}
438D - The Child and Sequence的更多相关文章
- 题解——CodeForces 438D The Child and Sequence
题面 D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input ...
- 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 ...
- Codeforces 438D The Child and Sequence
题意:给定一个n个数的序列,完成以下3个操作: 1.给定区间求和 2.给定区间对x取模 3.单点修改 对一个数取模,这个数至少折半.于是我们记一个最大值max,如果x>max则不做处理. #in ...
- 2018.07.23 codeforces 438D. The Child and Sequence(线段树)
传送门 线段树维护区间取模,单点修改,区间求和. 这题老套路了,对一个数来说,每次取模至少让它减少一半,这样每次单点修改对时间复杂度的贡献就是一个log" role="presen ...
- CF(438D) The Child and Sequence(线段树)
题意:对数列有三种操作: Print operation l, r. Picks should write down the value of . Modulo operation l, r, x. ...
- CodeForces 438D The Child and Sequence (线段树 暴力)
传送门 题目大意: 给你一个序列,要求在序列上维护三个操作: 1)区间求和 2)区间取模 3)单点修改 这里的操作二很讨厌,取模必须模到叶子节点上,否则跑出来肯定是错的.没有操作二就是线段树水题了. ...
- Codeforce 438D-The Child and Sequence 分类: Brush Mode 2014-10-06 20:20 102人阅读 评论(0) 收藏
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
- 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 ...
- Codeforces Round #250 (Div. 1) D. The Child and Sequence(线段树)
D. The Child and Sequence time limit per test 4 seconds memory limit per test 256 megabytes input st ...
随机推荐
- P2339 提交作业usaco(区间dp)
P2339 提交作业usaco 题目背景 usaco 题目描述 贝西在哞哞大学选修了 C 门课,她要把所有作业分别交给每门课的老师,然后去车站和同学们一起回家.每个老师在各自的办公室里,办公室要等他们 ...
- ROS-USB摄像头
前言:演示使用usb摄像头功能,推荐使用方法二. 首先要有一个usb摄像头,本次使用的是罗技(Logitech)摄像头. 一.使用软件库里的uvc-camera功能包 1.1 检查摄像头 lsusb ...
- spring controller接口中,用pojo对象接收页面传递的参数,发现spring在对pojo对象赋值时,有一定顺序的问题
1.我的项目中的实体类都继承了基类entityBase,里面封装了分页的一些属性,pageindex.pagesize.pagerownum等. 2.思路是页面可以灵活的传递分页参数,比如当前页pag ...
- 用Movie显示gif(1)SimpleGif
代码如下: import android.content.Context; import android.graphics.Canvas; import android.graphics.Movie; ...
- redis 配置多个ip 解决方案
因为在 redis 中bind 指定的ip 其实为同一网段或localhost 监听ip,在这里配置 内网其他网段或者外网多个ip 后 重启 redis 是不会成功的, 这边建议使用 折中方案,开通 ...
- log4j2异步日志解读(二)AsyncLogger
前文已经讲了log4j2的AsyncAppender的实现[log4j2异步日志解读(一)AsyncAppender],今天我们看看AsyncLogger的实现. 看了这个图,应该很清楚AsyncLo ...
- ajax怎么理解?
Ajix是创建交互式网页的前端网页开发技术,不是一种语言,ajax是基于http来传输数据的,他是利用浏览器提供操作http的接口(XMLHttpRequest或者activeXobject),来操作 ...
- R语言曲线拟合函数(绘图)
曲线拟合:(线性回归方法:lm) 1.x排序 2.求线性回归方程并赋予一个新变量 z=lm(y~x+I(x^2)+...) 3.plot(x,y) #做y对x的散点图 4.lines(x ...
- TI 77GHZ雷达开发套件 RDP-DC100
RDP-DC100用户使用手册 目录 1. 硬件说明... 3 1.1. 官方处理板的修 ...
- 腾讯云 LNMP+wordpress 搭建个人网站
折腾了好几个小时才弄好(php nginx略知一二),其实一点都不难! 以此记录一下,献给首次搭建的朋友们!! 1)准备工作:(因为个人用的ubuntu16.04 LTS系统 所以这是debian版 ...