Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".

The rules of this TV show are as follows: there are n

multisets numbered from 1 to n. Each of them is initially empty. Then, q

events happen; each of them is in one of the four possible types:

  • 1 x v — set the x

-th multiset to a singleton {v}

  • 2 x y z — set the x

-th multiset to a union of the y-th and the z-th multiset. For example: {1,3}∪{1,4,4}={1,1,3,4,4}

  • 3 x y z — set the x

-th multiset to a product of the y-th and the z-th multiset. The product A×B of two multisets A, B is defined as {gcd(a,b)∣a∈A,b∈B}, where gcd(p,q) is the greatest common divisor of p and q. For example: {2,2,3}×{1,4,6}={1,2,2,1,2,2,1,1,3}

  • 4 x v — the participant is asked how many times number v

occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2

  • only.

Note, that x

, y and z described above are not necessarily different. In events of types 2 and 3

, the sum or the product is computed first, and then the assignment is performed.

Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4

-th type?

Input

The first line contains two integers n

and q (1≤n≤105, 1≤q≤106

) — the number of multisets and the number of events.

Each of the following q

lines describes next event in the format given in statement. It's guaranteed that 1≤x,y,z≤n and 1≤v≤7000

always holds.

It's guaranteed that there will be at least one event of the 4

-th type.

Output

Print a string which consists of digits 0

and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4

-th type.

Example

Input
4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4
Output
010101

Note

Here is how the multisets look in the example test after each of the events; i

is the number of queries processed so far:

题意:

n个可重集,有Q次操作
   1 u v 表示将第u个可重集的元素置为1个v
   2 u a b 表示将第u个可重集置为第a个可重集和第b个可重集的并集
   3 u a b 表示将第u个可重集置为第a个可重集的每个元素和第b个可重集的每个元素的gcd的并集
   4 u v 表示求在第u个可重集中元素v的出现次数是奇数还是偶数
   n<=1e5 Q<=1e6 1<=v<=7000

思路:由于是只要求奇数还是偶数,我们整个过程只需要保存0和1即可,我们用莫比乌斯来求是否存在一个gcd,即保存当前集合是因子的奇偶性。那么对于2和3,我们可以直接操作(分别是^ &)了。

假设我们知道了因子的数量的奇偶性,假设保存在s[]里面。  vis[gcd]=mu(d/gcd)*s[d];所以对于每个gcd,我们预处理出mu(d/gcd)!=0的位置d,保存到b[]里面。

由于只求奇偶,1和-1的效果等效,结果和s[x]*b[y]的1的数量奇偶相同;

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
const int maxm=;
bitset<maxm>s[maxn],b[maxm];
int mu[maxm],p[maxm],cnt;bool vis[maxm];
vector<int>G[maxm];
void init()
{
mu[]=;
rep(i,,maxm-){
if(!vis[i]) p[++cnt]=i,mu[i]=-;
rep(j,,cnt){
if(i*p[j]>=maxm) break;
vis[i*p[j]]=;
if(!(i%p[j])) {mu[i*p[j]]=; break;}
mu[i*p[j]]=-mu[i];
}
}
rep(i,,maxm-)
for(int j=i,k=;j<=maxm-;j+=i,k++){
G[j].push_back(i);
if(mu[k]!=) b[i][j]=;
}
}
int main()
{
int N,M,opt,x,y,z;
init();
scanf("%d%d",&N,&M);
while(M--){
scanf("%d",&opt);
if(opt==){
scanf("%d%d",&x,&y);
s[x].reset();
rep(i,,G[y].size()-)
s[x][G[y][i]]=s[x][G[y][i]]^;
}
else if(opt==){
scanf("%d%d%d",&x,&y,&z);
s[x]=s[y]^s[z];
}
else if(opt==){
scanf("%d%d%d",&x,&y,&z);
s[x]=s[y]&s[z];
}
else {
scanf("%d%d",&x,&y);
if((s[x]&b[y]).count()&) putchar('');
else putchar('');
}
}
return ;
}

CodeForces - 1097F:Alex and a TV Show (bitset & 莫比乌斯容斥)的更多相关文章

  1. Codeforces 1097F Alex and a TV Show (莫比乌斯反演)

    题意:有n个可重集合,有四种操作: 1:把一个集合设置为单个元素v. 2:两个集合求并集. 3:两个集合中的元素两两求gcd,然后这些gcd形成一个集合. 4:问某个可重复集合的元素v的个数取模2之后 ...

  2. Codeforces 1097F. Alex and a TV Show

    传送门 由于只要考虑 $\mod 2$ 意义下的答案,所以我们只要维护一堆的 $01$ 容易想到用 $bitset$ 瞎搞...,发现当复杂度 $qv/32$ 是可以过的... 一开始容易想到对每个集 ...

  3. Codeforces Round #330 (Div. 2) B. Pasha and Phone 容斥定理

    B. Pasha and Phone Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/pr ...

  4. Codeforces Round #258 (Div. 2) E. Devu and Flowers 容斥

    E. Devu and Flowers 题目连接: http://codeforces.com/contest/451/problem/E Description Devu wants to deco ...

  5. Codeforces 1097 Alex and a TV Show

    传送门 除了操作 \(3\) 都可以 \(bitset\) 现在要维护 \[C_i=\sum_{gcd(j,k)=i}A_jB_k\] 类比 \(FWT\),只要求出 \(A'_i=\sum_{i|d ...

  6. Codeforces Round #428 (Div. 2) D. Winter is here 容斥

    D. Winter is here 题目连接: http://codeforces.com/contest/839/problem/D Description Winter is here at th ...

  7. codeforces 342D Xenia and Dominoes(状压dp+容斥)

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud D. Xenia and Dominoes Xenia likes puzzles ...

  8. CodeForces - 803F: Coprime Subsequences(莫比乌斯&容斥)

    Let's call a non-empty sequence of positive integers a1, a2... ak coprime if the greatest common div ...

  9. Codeforces Round #330 (Div. 2)B. Pasha and Phone 容斥

    B. Pasha and Phone   Pasha has recently bought a new phone jPager and started adding his friends' ph ...

随机推荐

  1. js 复制对象的深复制与浅复制

    1.潜复制(修改新对象会改变原对象) var baz = {a:'hello', b: {c:'my', d:'friend'}} var foo = baz foo.a="better&q ...

  2. 逆袭之旅DAY24.XIA.数组练习

    2018-07-20 08:40:19 1. public void stringSort(){ String[] s = new String[]{"George"," ...

  3. Entrust - Laravel 用户权限系统解决方案

    Zizaco/Entrust 是 Laravel 下 用户权限系统 的解决方案, 配合 用户身份认证 扩展包 Zizaco/confide 使用, 可以快速搭建出一套具备高扩展性的用户系统. Conf ...

  4. jsp下载excel文件

    jsp下载excel文件的的实现方法很多,今天也遇到这个问题,乱敲了一阵,终于搞定了,记下来和朋友们分享吧. 假设需要下载excel文件的jsp页面名为:down.jsp 对应的后台action名为: ...

  5. zookeeper集群环境搭建(纯zookeeper)

    1.首先在三台机子上放上zookeeper的解压包,解压. 然后的话zookeeper是依赖于jdk的,那么也应该安装jdk,这里不详细说明了. mv zookeeper-3.4.5 zookeepe ...

  6. Python Oracle连接与操作封装

    一.封装方式一 #encoding:utf-8 import cx_Oracleclass Oracle_Status_Output:    def __init__(self,db_name,db_ ...

  7. SQL-17 获取当前(to_date='9999-01-01')薪水第二多的员工的emp_no以及其对应的薪水salary

    题目描述 获取当前(to_date='9999-01-01')薪水第二多的员工的emp_no以及其对应的薪水salaryCREATE TABLE `salaries` (`emp_no` int(11 ...

  8. Linux(centos) 下curl模拟Http get / post请求 [ curl ]

    一.get请求 curl "http://www.baidu.com"  如果这里的URL指向的是一个文件或者一幅图都可以直接下载到本地 curl -i "http:// ...

  9. http协议相关

    HTTP请求方法 HTTP消息头 HTTP请求头 HTTP响应头 HTTP cookie机制和实现原理 HTTP请求方法 超文本传输协议(HTTP, HyperText Transfer Protoc ...

  10. codeforces983A(数学题)

    A. Finite or not? time limit per test 1 second memory limit per test 256 megabytes input standard in ...