B. Vasya and Books codeforce 1073B

B. Vasya and Books

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has got nn books, numbered from 11 to nn, arranged in a stack. The topmost book has number a1a1, the next one — a2a2, and so on. The book at the bottom of the stack has number anan. All numbers are distinct.

Vasya wants to move all the books to his backpack in nn steps. During ii-th step he wants to move the book number bibi into his backpack. If the book with number bibi is in the stack, he takes this book and all the books above the book bibi, and puts them into the backpack; otherwise he does nothing and begins the next step. For example, if books are arranged in the order [1,2,3][1,2,3] (book 11 is the topmost), and Vasya moves the books in the order [2,1,3][2,1,3], then during the first step he will move two books (11 and 22), during the second step he will do nothing (since book 11 is already in the backpack), and during the third step — one book (the book number 33). Note that b1,b2,…,bnb1,b2,…,bn are distinct.

Help Vasya! Tell him the number of books he will put into his backpack during each step.

Input

The first line contains one integer n (1≤n≤2⋅105)n (1≤n≤2⋅105) — the number of books in the stack.

The second line contains nn integers a1,a2,…,an (1≤ai≤n)a1,a2,…,an (1≤ai≤n) denoting the stack of books.

The third line contains nn integers b1,b2,…,bn (1≤bi≤n)b1,b2,…,bn (1≤bi≤n) denoting the steps Vasya is going to perform.

All numbers a1…ana1…an are distinct, the same goes for b1…bnb1…bn.

Output

Print nn integers. The ii-th of them should be equal to the number of books Vasya moves to his backpack during the ii-th step.

Examples

input

Copy

3
1 2 3
2 1 3

output

Copy

2 0 1 

input

Copy

5
3 1 4 2 5
4 5 1 3 2

output

Copy

3 2 0 0 0 

input

Copy

6
6 5 4 3 2 1
6 5 3 4 2 1

output

Copy

1 1 2 0 1 1 

Note

The first example is described in the statement.

In the second example, during the first step Vasya will move the books [3,1,4][3,1,4]. After that only books 22 and 55 remain in the stack (22 is above 55). During the second step Vasya will take the books 22 and 55. After that the stack becomes empty, so during next steps Vasya won't move any books.

暴力用队列果断超时,然后就想用三个数组模拟

#include<iostream>
using namespace std;
int a[200005];
int b[200005];
int pos[200005];
int main(){
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>a[i];
		pos[a[i]]=i;
	}
	for(int i=0;i<n;i++){
		cin>>b[i];
	}
	int q=-1;
	for(int i=0;i<n;i++){
		if(q<pos[b[i]]){
			if(!i)
			cout<<pos[b[i]]-q;
			else
			cout<<" "<<pos[b[i]]-q;
			q=pos[b[i]];
		}else{
			cout<<" "<<0;
		}
	}
	cout<<endl;
	return 0;
}

 

### 关于 Vasya 和多重集的编程竞赛算法题目解决方案 #### 题目描述 给定一个多重集合 `s`,目标是将其分割成两个新的多重集合 `a` 和 `b` (其中一个可以为空),使得这两个新集合中的“好数”的数量相等。“好数”定义为在一个特定多集中恰好只出现一次的数字。 为了实现这一目标,需要考虑如何有效地统计并分配这些元素到不同的子集中去[^1]。 #### 解决思路 一种有效的解决方法是从输入数据的特点出发思考。如果某个数值在整个原始集合中出现了偶数次,则该值可以在不影响最终结果的情况下被平均分入两个子集中;而对于那些仅出现奇数次数的情况,则必须小心处理以确保能够达成平衡条件——即让尽可能多的不同类型的单例项分别进入各自的目标组内[^2]。 具体来说: - 对于任何频率大于等于两次(无论是奇还是偶)的数据点而言,总是能通过适当划分来满足上述要求; - 当遇到频度为一的情形时,就需要额外注意了:因为这直接影响着能否成功创建具有相同数目唯一成员的新分区。 因此,在实际编码过程中应该优先处理那些重复率较高的项目,并记录下所有独一无二实例的位置以便后续操作使用。 #### Python 实现代码示例 下面是一个基于此逻辑编写的Python函数,用于求解这个问题: ```python from collections import Counter def can_split_equally(s): count = Counter(s) # 统计各元素出现次数 singletons = sum(1 for v in count.values() if v == 1) return singletons % 2 == 0 # 测试用例 test_cases = [ [1, 2, 2, 3], # True 可以分成 {1} 和 {2, 2, 3} [1, 2, 3, 4, 5], # False 单独存在的数字有五个无法平分 ] for case in test_cases: print(f"Input: {case}, Can Split Equally? :{can_split_equally(case)}") ``` 这个程序首先利用 `collections.Counter` 来计算每个整数在列表里边出现过的总次数。接着它会遍历所有的键值对,累积起所有只出现过一次(也就是所谓的 “好数” 或者说是单一实例)的数量。最后一步就是判断这样的特殊案例是不是构成了一个偶数序列长度 —— 如果是的话就意味着存在至少一组可行解;反之则不存在这样的一分为二方式。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值