Hdu 4578 Transformation （线段树 分类分析）
Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 10082 Accepted Submission(s): 2609
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
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.
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0
7489
题解
这道题有三种询问：set , add , mul。所以lazy标记要有三个，如果三个标记同时出现的处理方法——当更新set操作时，就把add标记和mul标记全部取消；当更新mul操作时，如果当前节点add标记存在，就把add标记改为：add * mul。这样的话就可以在PushDown()操作中先执行set，然后mul，最后add。
C++代码一
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = ;
#define left v<<1
#define right v<<1|1
#define mod 10007
struct node
{
int l ,r , value ;
int eq , add , mul ;
}tree[maxn<<];
void build(int l , int r , int v)
{
tree[v].l = l ;
tree[v].r = r ;
tree[v].add = ; tree[v].mul = ;tree[v].eq = - ;
if(l == r)
{tree[v].eq = ; return ;}
int mid = (l + r) >> ;
build(l , mid , left) ;
build(mid + , r , right) ;
}
void push_down(int v)
{
if(tree[v].l == tree[v].r)return ;
if(tree[v].eq != -)
{
tree[left].eq = tree[right].eq = tree[v].eq ;
tree[left].add = tree[right].add = ;
tree[left].mul = tree[right].mul = ;
tree[v].eq = -;
return ;
}
if(tree[v].mul != )
{
if(tree[left].eq != -)
tree[left].eq = (tree[left].eq*tree[v].mul)%mod ;
else
{
push_down(left) ;
tree[left].mul = (tree[left].mul*tree[v].mul)%mod ;
}
if(tree[right].eq != -)
tree[right].eq = (tree[right].eq*tree[v].mul)%mod ;
else
{
push_down(right) ;
tree[right].mul = (tree[right].mul*tree[v].mul)%mod ;
}
tree[v].mul = ;
}
if(tree[v].add)
{
if(tree[left].eq != -)
tree[left].eq = (tree[left].eq + tree[v].add)%mod ;
else
{
push_down(left) ;
tree[left].add = (tree[left].add + tree[v].add)%mod ;
}
if(tree[right].eq != -)
tree[right].eq = (tree[right].eq + tree[v].add)%mod ;
else
{
push_down(right) ;
tree[right].add = (tree[right].add + tree[v].add)%mod ;
}
tree[v].add = ;
}
}
void update(int l , int r , int v , int op , int c)
{
if(l <= tree[v].l && tree[v].r <= r)
{
if(op == )
{
tree[v].add = ;tree[v].mul = ;
tree[v].eq = c ;
return ;
}
if(tree[v].eq != -)
{
if(op == )tree[v].eq = (tree[v].eq + c)%mod ;
else tree[v].eq = (tree[v].eq*c)%mod ;
}
else
{
push_down(v) ;
if(op == )tree[v].add = (tree[v].add + c)%mod ;
else tree[v].mul = (tree[v].mul*c)%mod ;
}
return ;
}
push_down(v) ;
int mid = (tree[v].l + tree[v].r) >> ;
if(l <= mid)update(l , r ,left , op , c) ;
if(r > mid)update(l , r , right , op , c) ;
}
int query(int l , int r , int v , int q)
{
if(tree[v].l >= l && tree[v].r <= r && tree[v].eq != -)
{
int ans = ;
for(int i = ;i <= q;i++)
ans = (ans * tree[v].eq)%mod ;
return (ans*((tree[v].r - tree[v].l + )%mod))%mod ;
}
push_down(v) ;
int mid = (tree[v].l + tree[v].r) >> ;
if(l > mid)return query(l , r , right, q) ;
else if(r <= mid)return query(l , r ,left ,q) ;
else return (query(l , mid , left , q) + query(mid + , r , right , q))%mod ;
}
int main()
{
//freopen("in.txt" ,"r" , stdin) ;
int n , m ;
while(scanf("%d%d" , &n , &m) &&(n+m))
{
int op , x , y , c;
build( , n , ) ;
while(m--)
{
scanf("%d%d%d%d" , &op , &x , &y , &c) ;
if(op == )
printf("%d\n" , (query(x, y , , c)%mod)) ;
else update(x , y , , op , c) ;
}
}
return ;
}
C++代码二
解释
平方和这样来推：(a + c)2 = a2 + c2 + 2ac , 即sum2[rt] = sum2[rt] + (r - l + 1) * c * c + 2 * sum1[rt] * c；
立方和这样推：(a + c)3 = a3 + c3 + 3a(a2 + ac) , 即sum3[rt] = sum3[rt] + (r - l + 1) * c * c * c + 3 * c * (sum2[rt] + sum1[rt] * c)；
几个注意点：add标记取消的时候是置0，mul标记取消的时候是置1；在PushDown()中也也要注意取消标记，如set操作中取消add和mul，mul操作中更新add； 在add操作中要注意sum3 , sum2 , sum1的先后顺序，一定是先sum3 , 然后sum2 , 最后sum1； int容易爆，还是用LL要保险一点； 最后就是运算较多，不要漏掉东西。
当然这种方法有取巧的成分
#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <string>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL __int64
typedef long long ll;
#define eps 1e-8
#define INF INT_MAX
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int MOD = ;
const int maxn = + ;
const int N = ;
ll add[maxn << ] , set[maxn << ] , mul[maxn << ];
ll sum1[maxn << ] , sum2[maxn << ] , sum3[maxn << ];
void PushUp(int rt)
{
sum1[rt] = (sum1[rt << ] + sum1[rt << | ]) % MOD;
sum2[rt] = (sum2[rt << ] + sum2[rt << | ]) % MOD;
sum3[rt] = (sum3[rt << ] + sum3[rt << | ]) % MOD;
}
void build(int l , int r , int rt)
{
add[rt] = set[rt] = ;
mul[rt] = ;
if(l == r) {
sum1[rt] = sum2[rt] = sum3[rt] = ;
return;
}
int m = (l + r) >> ;
build(lson);
build(rson);
PushUp(rt);
}
void PushDown(int rt , int len)
{
if(set[rt]) {
set[rt << ] = set[rt << | ] = set[rt];
add[rt << ] = add[rt << | ] = ; //注意这个也要下放
mul[rt << ] = mul[rt << | ] = ;
ll tmp = ((set[rt] * set[rt]) % MOD) * set[rt] % MOD;
sum1[rt << ] = ((len - (len >> )) % MOD) * (set[rt] % MOD) % MOD;
sum1[rt << | ] = ((len >> ) % MOD) * (set[rt] % MOD) % MOD;
sum2[rt << ] = ((len - (len >> )) % MOD) * ((set[rt] * set[rt]) % MOD) % MOD;
sum2[rt << | ] = ((len >> ) % MOD) * ((set[rt] * set[rt]) % MOD) % MOD;
sum3[rt << ] = ((len - (len >> )) % MOD) * tmp % MOD;
sum3[rt << | ] = ((len >> ) % MOD) * tmp % MOD;
set[rt] = ;
}
if(mul[rt] != ) { //这个就是mul[rt] != 1 ， 当时我这里没注意所以TLE了
mul[rt << ] = (mul[rt << ] * mul[rt]) % MOD;
mul[rt << | ] = (mul[rt << | ] * mul[rt]) % MOD;
if(add[rt << ]) //注意这个也要下放
add[rt << ] = (add[rt << ] * mul[rt]) % MOD;
if(add[rt << | ])
add[rt << | ] = (add[rt << | ] * mul[rt]) % MOD;
ll tmp = (((mul[rt] * mul[rt]) % MOD * mul[rt]) % MOD);
sum1[rt << ] = (sum1[rt << ] * mul[rt]) % MOD;
sum1[rt << | ] = (sum1[rt << | ] * mul[rt]) % MOD;
sum2[rt << ] = (sum2[rt << ] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD;
sum2[rt << | ] = (sum2[rt << | ] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD;
sum3[rt << ] = (sum3[rt << ] % MOD) * tmp % MOD;
sum3[rt << | ] = (sum3[rt << | ] % MOD) * tmp % MOD;
mul[rt] = ;
}
if(add[rt]) {
add[rt << ] += add[rt]; //add是+= ， mul是*=
add[rt << | ] += add[rt];
ll tmp = (add[rt] * add[rt] % MOD) * add[rt] % MOD; //注意sum3 , sum2 , sum1的先后顺序
sum3[rt << ] = (sum3[rt << ] + (tmp * (len - (len >> )) % MOD) + * add[rt] * ((sum2[rt << ] + sum1[rt << ] * add[rt]) % MOD)) % MOD;
sum3[rt << | ] = (sum3[rt << | ] + (tmp * (len >> ) % MOD) + * add[rt] * ((sum2[rt << | ] + sum1[rt << | ] * add[rt]) % MOD)) % MOD;
sum2[rt << ] = (sum2[rt << ] + ((add[rt] * add[rt] % MOD) * (len - (len >> )) % MOD) + ( * sum1[rt << ] * add[rt] % MOD)) % MOD;
sum2[rt << | ] = (sum2[rt << | ] + (((add[rt] * add[rt] % MOD) * (len >> )) % MOD) + ( * sum1[rt << | ] * add[rt] % MOD)) % MOD;
sum1[rt << ] = (sum1[rt << ] + (len - (len >> )) * add[rt]) % MOD;
sum1[rt << | ] = (sum1[rt << | ] + (len >> ) * add[rt]) % MOD;
add[rt] = ;
}
}
void update(int L , int R , int c , int ch , int l , int r , int rt)
{
if(L <= l && R >= r) {
if(ch == ) {
set[rt] = c;
add[rt] = ;
mul[rt] = ;
sum1[rt] = ((r - l + ) * c) % MOD;
sum2[rt] = ((r - l + ) * ((c * c) % MOD)) % MOD;
sum3[rt] = ((r - l + ) * (((c * c) % MOD) * c % MOD)) % MOD;
} else if(ch == ) {
mul[rt] = (mul[rt] * c) % MOD;
if(add[rt])
add[rt] = (add[rt] * c) % MOD;
sum1[rt] = (sum1[rt] * c) % MOD;
sum2[rt] = (sum2[rt] * (c * c % MOD)) % MOD;
sum3[rt] = (sum3[rt] * ((c * c % MOD) * c % MOD)) % MOD;
} else if(ch == ) {
add[rt] += c;
ll tmp = (((c * c) % MOD * c) % MOD * (r - l + )) % MOD; //(r - l + 1) * c^3
sum3[rt] = (sum3[rt] + tmp + * c * ((sum2[rt] + sum1[rt] * c) % MOD)) % MOD;
sum2[rt] = (sum2[rt] + (c * c % MOD * (r - l + ) % MOD) + * sum1[rt] * c) % MOD;
sum1[rt] = (sum1[rt] + (r - l + ) * c) % MOD;
}
return;
}
PushDown(rt , r - l + );
int m = (l + r) >> ;
if(L > m)
update(L , R , c , ch , rson);
else if(R <= m)
update(L , R , c , ch , lson);
else {
update(L , R , c , ch , lson);
update(L , R , c , ch , rson);
}
PushUp(rt);
}
ll query(int L , int R , int p , int l , int r , int rt)
{
if(L <= l && R >= r) {
if(p == )
return sum1[rt] % MOD;
else if(p == )
return sum2[rt] % MOD;
else
return sum3[rt] % MOD;
}
PushDown(rt , r - l + );
int m = (l + r) >> ;
if(L > m)
return query(L , R , p , rson);
else if(R <= m)
return query(L , R , p , lson);
else
return (query(L , R , p , lson) + query(L , R , p , rson)) % MOD;
}
int main()
{
int n , m;
int a , b , c , ch;
while(~scanf("%d %d" , &n , &m))
{
if(n == && m == )
break;
build( , n , );
while(m--) {
scanf("%d %d %d %d" , &ch , &a , &b , &c);
if(ch != ) {
update(a , b , c , ch , , n , );
} else {
printf("%lld\n" , query(a , b , c , , n , ));
}
}
}
return ;
}
Hdu 4578 Transformation （线段树 分类分析）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（线段树区间修改）
https://cn.vjudge.net/problem/HDU-4578 题意 4种操作,区间加,区间乘,区间变为一个数,求区间的和.平方和以及立方和. 分析 明显线段树,不过很麻烦..看kuan ...
随机推荐
- node.js从入门到起飞
第一个node程序: 首先创建一个js文件,命名index.js(可随意),然后在文件里面输入 : console.log("Hello World"); 使用 Git Bash ...
- 微信小程序-tabBar-注意事项
tabBar.list[0].selectedIconPath 文件格式错误,仅支持 .png..jpg..jpeg 格式
- computed属性和watcher
computed属性 在模板中使用表达式是非常方便直接的,然而这只适用于简单的操作.在模板中放入太多的逻辑,会使模板过度膨胀和难以维护.例如: <div id="example&quo ...
- mysql DEFAULT约束 语法
mysql DEFAULT约束 语法 作用:用于向列中插入默认值. 说明:如果没有规定其他的值,那么会将默认值添加到所有的新记录.直线电机 mysql DEFAULT约束 示例 //在 "P ...
- Ubuntu redis 实战 持久化策略 主从复制 以及 故障恢复
推荐文章 redis数据结构学习 redis持久化 redis主从复制 redis哨兵
- （22）Python练习项目集
文本操作 逆转字符串——输入一个字符串,将其逆转并输出. 拉丁猪文字游戏——这是一个英语语言游戏.基本规则是将一个英语单词的第一个辅音音素的字母移动到词尾并且加上后缀-ay(譬如“banana”会变成 ...
- Spring MVC Captcha 验证码
本文专为不想一遍遍手写验证码代码的”懒猿“而生,只需要添加部分配置文件,拷贝一个控制层代码,就能在页面中方便使用.话不多说,进入主题: 一. Spring MVC 环境实现方式: 1.1 spring ...
- iOS Beta 升级或降级
https://sspai.com/post/45442 public beta网站上安装profile, 去手机看更新 developer beta, 登录开发者网站, downloads, 下载p ...
- you have not created a boot efi partition
rhel6.8 自定义分区,创建 /boot/efi 分区找不到”EFI System Partition“文件系统, 选其他文件系统 next时都提示:you have not created a ...
- What is the difference between Kill and Kill -9 command in Unix?
w difference kill -9 pid and kill pid command - Ask Ubuntu https://askubuntu.com/questions/791841/d ...