Boring Game

1071 - Boring Game

Time Limit：1s Memory Limit：1024MByte

Submissions：541Solved：155

DESCRIPTION

In a mysterious cave, PigVan ( Mr.Van's pet ) has lived for thousands years. PigVan absorbed the power of nature, and it may pretend a human in speaking, walking and so on.

One day, he bought some valuable stone, and divided them into n piles of stone where thei^th pile (1≤i≤n) contains valuesa_i .

After PigVan put them in a line, he wants to play a game.

In the boring game, he can do this operation:

Choose a stone pile a_i （i>1）and its two adjacent piles a_i-1, a_i+1, turn(a_i-1, a_i, a_i+1) to(a_i-1 + a_i, -a_i, a_i + a_i+1).

PigVan wonders whether he can get (b₁, b₂, b₃, …, b_n) after several operations.

Note:

If you choose the last pile a_n, the operation will be( a_n-1 + a_n, -a_n ) .

INPUT

The first line is a single integer $T$, indicating the number of test cases.
For each test case:
In the first line, there are only one integer $n$(n≤10⁵), indicating the number of food piles.
The second line is $n$ integers indicate sequence $a$ ( | a_i | ≤ 10⁶).
The third line is $n$ integers indicate sequence $b$ ( | b_i | ≤ 10⁶).

OUTPUT

For each test case, just print 'Yes' if PigVan can get $b$ after some operations; otherwise, print 'No'.

SAMPLE INPUT

2

6

1 6 9 4 2 0

7 -6 19 2 -6 6

4

1 2 3 4 

4 2 1 3

SAMPLE OUTPUT

Yes

No

题意：

考虑一次操作
(ai-1, ai, ai+1)->(ai-1 + ai, -ai, ai+1 + ai)
如果考虑前缀和，那么一次操作等效为
(si-1, si, si+1)->(si, si-1, si+1)
即对于前缀和而言，他只是交换了位置。
因此，我们只需求出前缀和，然后看元素是否对等就行了。

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>using namespace std;int a[1000000],b[1000000];int c[1000000],d[1000000];int main(){    int t;    scanf("%d",&t);    while(t--)    {       int n;       memset(c,0,sizeof(c));       memset(d,0,sizeof(d));       scanf("%d",&n);       for(int i=1;i<=n;i++)       {           scanf("%d",&a[i]);           c[i]=c[i-1]+a[i];       }       for(int i=1;i<=n;i++)       {           scanf("%d",&b[i]);           d[i]=d[i-1]+b[i];       }       sort(c+1,c+n+1);       sort(d+1,d+n+1);       int sum=0;       for(int i=1;i<=n;i++)       {           if(c[i]==d[i])           {               sum++;           }       }       if(sum==n)       {           printf("Yes\n");       }       else        printf("No\n");    }}