D. The Child and Sequence
time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

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:

  1. Print operation l, r. Picks should write down the value of .
  2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[imod x for each i (l ≤ i ≤ r).
  3. 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?

Input

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.
Output

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

#include <bits/stdc++.h>
;
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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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. ...

  6. 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. ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. bzoj 3534: [Sdoi2014]重建【矩阵树定理】

    啊啊啊无脑背过果然不可取 比如这道题就不会写 参考:https://blog.csdn.net/iamzky/article/details/41317333 #include<iostream ...

  2. bzoj4720: [Noip2016]换教室(期望dp)

    4720: [Noip2016]换教室 Time Limit: 20 Sec  Memory Limit: 512 MBSubmit: 1294  Solved: 698[Submit][Status ...

  3. [Swift]Set(集)转换为Array(数组)

    ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs. ...

  4. python自动化测试学习笔记-10YAML

    之前学习的编写测试用例的方法,都是从excel中编写接口的测试用例,然后通过读取excel文件进行接口自动化测试,这种方式我们叫做数据驱动的方式,由于excel操作起来不灵活,无法实现关联关系的接口测 ...

  5. Spring-security配置代码

    @Configuration public static class WebSecurityConfigurer extends WebSecurityConfigurerAdapter{ @Over ...

  6. H5活动的一些事

    ISUX团队镇楼:https://isux.tencent.com/nine-question-of-swipe-html5-page.html IE6.7.8支持html5新元素 : http:// ...

  7. [ CodeForces 438 D ] The Child and Sequence

    \(\\\) \(Description\) 维护长为 \(N\) 的数列,\(M\)次操作,支持单点修改,区间取模,查询区间和. \(N,M\le 10^5\) \(\\\) \(Solution\ ...

  8. 42使用NanoPiM1Plus在Android4.4.2下的录音测试

    42使用NanoPiM1Plus在Android4.4.2下的录音测试 大文实验室/大文哥壹捌陆捌零陆捌捌陆捌贰21504965 AT qq.com完成时间:2017/12/5 17:51版本:V1. ...

  9. D1-mini esp8266的资料备份

    需要更新esp8266库 http://arduino.esp8266.com/stable/package_esp8266com_index.json 下载好即可.

  10. 自定义封装 banner 组件

    1. 效果图预览 2.基本功能 一个简单方便的轮播图组件,基于viewpager 基础上进行的封装.可设置 项目中图片,网络图片, View:支持循环自动播放,手势滑动切换,item点击事件,可设置 ...