CF1110E.Magic Stones

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题目描述

Grigory has nn magic stones, conveniently numbered from 11 to nn . The charge of the ii -th stone is equal to cic_i .

Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index ii , where 2in12 \le i \le n - 1 ), and after that synchronizes it with neighboring stones. After that, the chosen stone loses its own charge, but acquires the charges from neighboring stones. In other words, its charge cic_i changes to ci=ci+1+ci1cic_i' = c_{i + 1} + c_{i - 1} - c_i .

Andrew, Grigory's friend, also has nn stones with charges tit_i . Grigory is curious, whether there exists a sequence of zero or more synchronization operations, which transforms charges of Grigory's stones into charges of corresponding Andrew's stones, that is, changes cic_i into tit_i for all ii ?

输入格式

The first line contains one integer nn ( 2n1052 \le n \le 10^5 ) — the number of magic stones.

The second line contains integers c1,c2,,cnc_1, c_2, \ldots, c_n ( 0ci21090 \le c_i \le 2 \cdot 10^9 ) — the charges of Grigory's stones.

The second line contains integers t1,t2,,tnt_1, t_2, \ldots, t_n ( 0ti21090 \le t_i \le 2 \cdot 10^9 ) — the charges of Andrew's stones.

输出格式

If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print "Yes".

Otherwise, print "No".

输入输出样例

  • 输入#1

    4
    7 2 4 12
    7 15 10 12
    

    输出#1

    Yes
    
  • 输入#2

    3
    4 4 4
    1 2 3
    

    输出#2

    No
    

说明/提示

In the first example, we can perform the following synchronizations ( 11 -indexed):

  • First, synchronize the third stone [7,2,4,12][7,2,10,12][7, 2, \mathbf{4}, 12] \rightarrow [7, 2, \mathbf{10}, 12] .
  • Then synchronize the second stone: [7,2,10,12][7,15,10,12][7, \mathbf{2}, 10, 12] \rightarrow [7, \mathbf{15}, 10, 12] .

In the second example, any operation with the second stone will not change its charge.

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