Transformation

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 49 Accepted Submission(s): 16

Problem Description

Yuanfang is puzzled with the question below:
There are n integers, a₁, a₂, …, a_n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a_x and a_y to c, inclusive. In other words, do transformation a_k<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a_x and a_y inclusive. In other words, get the result of a_x^p+a_x+1^p+…+a_y ^p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.

Input

There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

Sample Input

5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output

307
7489

Source

2013ACM-ICPC杭州赛区全国邀请赛

很裸的线段树的题目。

但是做起来比较麻烦。

我用sum1,sum2,sum3分别代表和、平方和、立方和。

懒惰标记使用三个变量：

lazy1:是加的数

lazy2:是乘的倍数

lazy3:是赋值为一个常数，为0表示没有。

更新操作需要注意很多细节。

 /* **********************************************

 Author      : kuangbin

 Created Time: 2013/8/10 13:24:03

 File Name   : F:\2013ACM练习\比赛练习\2013杭州邀请赛重现\1003.cpp

 *********************************************** */

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <set>

 #include <map>

 #include <string>

 #include <math.h>

 #include <stdlib.h>

 using namespace std;

 const int MOD = ;

 const int MAXN = ;

 struct Node

 {

     int l,r;

     int sum1,sum2,sum3;

     int lazy1,lazy2,lazy3;

 }segTree[MAXN*];

 void build(int i,int l,int r)

 {

     segTree[i].l = l;

     segTree[i].r = r;

     segTree[i].sum1 = segTree[i].sum2 = segTree[i].sum3 = ;

     segTree[i].lazy1 = segTree[i].lazy3 = ;

     segTree[i].lazy2 = ;

     int mid = (l+r)/;

     if(l == r)return;

     build(i<<,l,mid);

     build((i<<)|,mid+,r);

 }

 void push_up(int i)

 {

     if(segTree[i].l == segTree[i].r)

         return;

     segTree[i].sum1 = (segTree[i<<].sum1 + segTree[(i<<)|].sum1)%MOD;

     segTree[i].sum2 = (segTree[i<<].sum2 + segTree[(i<<)|].sum2)%MOD;

     segTree[i].sum3 = (segTree[i<<].sum3 + segTree[(i<<)|].sum3)%MOD;

 }

 void push_down(int i)

 {

     if(segTree[i].l == segTree[i].r) return;

     if(segTree[i].lazy3 != )

     {

         segTree[i<<].lazy3 = segTree[(i<<)|].lazy3 = segTree[i].lazy3;

         segTree[i<<].lazy1 = segTree[(i<<)|].lazy1 = ;

         segTree[i<<].lazy2 = segTree[(i<<)|].lazy2 = ;

         segTree[i<<].sum1 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD;

         segTree[i<<].sum2 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;

         segTree[i<<].sum3 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;

         segTree[(i<<)|].sum1 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD;

         segTree[(i<<)|].sum2 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;

         segTree[(i<<)|].sum3 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;

         segTree[i].lazy3 = ;

     }

     if(segTree[i].lazy1 !=  || segTree[i].lazy2 != )

     {

         segTree[i<<].lazy1 = ( segTree[i].lazy2*segTree[i<<].lazy1%MOD + segTree[i].lazy1 )%MOD;

         segTree[i<<].lazy2 = segTree[i<<].lazy2*segTree[i].lazy2%MOD;

         int sum1,sum2,sum3;

         sum1 = (segTree[i<<].sum1*segTree[i].lazy2%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD)%MOD;

         sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i<<].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[i<<].sum1%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;

         sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i<<].sum3 % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum2) % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum1) % MOD;

         sum3 = (sum3 + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;

         segTree[i<<].sum1 = sum1;

         segTree[i<<].sum2 = sum2;

         segTree[i<<].sum3 = sum3;

         segTree[(i<<)|].lazy1 = ( segTree[i].lazy2*segTree[(i<<)|].lazy1%MOD + segTree[i].lazy1 )%MOD;

         segTree[(i<<)|].lazy2 = segTree[(i<<)|].lazy2 * segTree[i].lazy2 % MOD;

         sum1 = (segTree[(i<<)|].sum1*segTree[i].lazy2%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD)%MOD;

         sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[(i<<)|].sum1%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;

         sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum3 % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum2) % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum1) % MOD;

         sum3 = (sum3 + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;

         segTree[(i<<)|].sum1 = sum1;

         segTree[(i<<)|].sum2 = sum2;

         segTree[(i<<)|].sum3 = sum3;

         segTree[i].lazy1 = ;

         segTree[i].lazy2 = ;

     }

 }

 void update(int i,int l,int r,int type,int c)

 {

     if(segTree[i].l == l && segTree[i].r == r)

     {

         c %= MOD;

         if(type == )

         {

             segTree[i].lazy1 += c;

             segTree[i].lazy1 %= MOD;

             segTree[i].sum3 = (segTree[i].sum3 + *segTree[i].sum2%MOD*c%MOD + *segTree[i].sum1%MOD*c%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD*c%MOD)%MOD;

             segTree[i].sum2 = (segTree[i].sum2 + *segTree[i].sum1%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD)%MOD;

             segTree[i].sum1 = (segTree[i].sum1 + (segTree[i].r - segTree[i].l + )*c%MOD)%MOD;

         }

         else if(type == )

         {

             segTree[i].lazy1 = segTree[i].lazy1*c%MOD;

             segTree[i].lazy2 = segTree[i].lazy2*c%MOD;

             segTree[i].sum1 = segTree[i].sum1*c%MOD;

             segTree[i].sum2 = segTree[i].sum2*c%MOD*c%MOD;

             segTree[i].sum3 = segTree[i].sum3*c%MOD*c%MOD*c%MOD;

         }

         else

         {

             segTree[i].lazy1 = ;

             segTree[i].lazy2 = ;

             segTree[i].lazy3 = c%MOD;

             segTree[i].sum1 = c*(segTree[i].r - segTree[i].l + )%MOD;

             segTree[i].sum2 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD;

             segTree[i].sum3 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD*c%MOD;

         }

         return;

     }

     push_down(i);

     int mid = (segTree[i].l + segTree[i].r)/;

     if(r <= mid)update(i<<,l,r,type,c);

     else if(l > mid)update((i<<)|,l,r,type,c);

     else

     {

         update(i<<,l,mid,type,c);

         update((i<<)|,mid+,r,type,c);

     }

     push_up(i);

 }

 int query(int i,int l,int r,int p)

 {

     if(segTree[i].l == l && segTree[i].r == r)

     {

         if(p == )return segTree[i].sum1;

         else if(p== )return segTree[i].sum2;

         else return segTree[i].sum3;

     }

     push_down(i);

     int mid = (segTree[i].l + segTree[i].r )/;

     if(r <= mid)return query(i<<,l,r,p);

     else if(l > mid)return query((i<<)|,l,r,p);

     else return (query(i<<,l,mid,p)+query((i<<)|,mid+,r,p))%MOD;

 }

 int main()

 {

     //freopen("in.txt","r",stdin);

     //freopen("out.txt","w",stdout);

     int n,m;

     while(scanf("%d%d",&n,&m) == )

     {

         if(n ==  && m == )break;

         build(,,n);

         int type,x,y,c;

         while(m--)

         {

             scanf("%d%d%d%d",&type,&x,&y,&c);

             if(type == )printf("%d\n",query(,x,y,c));

             else update(,x,y,type,c);

         }

     }

     return ;

 }

HDU 4578 Transformation （线段树）的更多相关文章

HDU 4578 Transformation --线段树，好题
题意: 给一个序列,初始全为0,然后有4种操作: 1. 给区间[L,R]所有值+c 2.给区间[L,R]所有值乘c 3.设置区间[L,R]所有值为c 4.查询[L,R]的p次方和(1<=p< ...
hdu 4578 Transformation 线段树
没什么说的裸线段树,注意细节就好了!!! 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> ...
hdu 4578 Transformation 线段树多种操作裸题
自己写了一个带结构体的WA了7.8次但是测了几组小数据都对..感觉问题应该出在模运算那里.写完这波题解去对拍一下. 以后线段树绝不写struct!一般的struct都带上l,r 但是一条线段的长度确 ...
Transformation HDU - 4578（线段树——懒惰标记的妙用）
Yuanfang is puzzled with the question below: There are n integers, a 1, a 2, …, a n. The initial val ...
hdu 4031 attack 线段树区间更新
Attack Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Subm ...
hdu 4288 离线线段树+间隔求和
Coder Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Su ...
hdu 3016 dp+线段树
Man Down Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total S ...
HDU 4578 - Transformation - [加强版线段树]
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4578 Problem Description Yuanfang is puzzled with the ...
HDU 4578——Transformation——————【线段树区间操作、确定操作顺序】
Transformation Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)T ...

随机推荐

Bit banging
Bit banging Bit banging is a technique for serial communications using software instead of dedicated ...
2017多校第7场 HDU 6127 Hard challenge 极角排序，双指针
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6127 题意:平面直角坐标系上有n个整点,第i个点有一个点权val,坐标为(xi,yi),其中不存在任 ...
Hierarchical Attention Based Semi-supervised Network Representation Learning
Hierarchical Attention Based Semi-supervised Network Representation Learning 1. 任务给定:节点信息网络目标:为每个节 ...
mybatis 一级缓存和二级缓存
1.默认是会话期内一级session缓存 2.二级缓存: 引入二级缓存的jar, 配置 ehcache.xml, mapper.xml引入缓存<cache type="org.myb ...
Leetcode 之Same Tree（48）
用递归比较简单,这里用迭代的方式实现.注意什么时候返回true,什么时候返回false. bool isSameTree(TreeNode *p, TreeNode *q) { stack<Tr ...
Leetcode 之Binary Tree Preorder Traversal（42）
树的先序遍历.定义一个栈,先压入中间结点并访问,然后依次压入右.左结点并访问. vector<int> preorderTraversal(TreeNode *root) { vector ...
linux下不解包查看tar包文件内容
为减少日志文件占用的空间,很多情况下我们会将日志文件以天或周为周期打包成tar.gz 包保存.虽然这样做有利空间充分利用,但当我们想查看压缩包内的内容时确很不方便.如果只是一个tar.gz文件,可以将 ...
C# For Bot Framework
Bot Framework是一个聊天机器人的框架,背后是微软的SDK,它可以使用C#和Nodejs开发,今天我尝试用创建一个比较简单Bot 参考地址:https://docs.microsoft.co ...
KMS使用CLion作为IDE来调试
KMS使用CLion作为IDE来调试,打开kms相应模块的目录,CLion自动识别相应的CMakeLists.txt. 然而会make失败,经搜索,看到Clion使用的自带的cmake,因此系统中安装 ...
Java实现蛇形矩阵
public class Solution { //下x++ 左y-- 上x-- 右y++ public void prints(int n) { int[][] mp = new int[n][n] ...

HDU 4578 Transformation （线段树）

Transformation

HDU 4578 Transformation （线段树）的更多相关文章

随机推荐

热门专题