Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set {−1,0,1}.
Anton can perform the following sequence of operations any number of times:
Choose any pair of indexes (i,j) such that 1≤i<j≤n. It is possible to choose the same pair (i,j) more than once.
Add ai to aj. In other words, j-th element of the array becomes equal to ai+aj.
For example, if you are given array [1,−1,0], you can transform it only to [1,−1,−1], [1,0,0] and [1,−1,1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1≤t≤10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1≤n≤105) — the length of arrays.
The second line of each test case contains n integers a1,a2,…,an (−1≤ai≤1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b1,b2,…,bn (−109≤bi≤109) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn’t exceed 105.
Output
For each test case, output one line containing “YES” if it’s possible to make arrays a and b equal by performing the described operations, or “NO” if it’s impossible.
You can print each letter in any case (upper or lower).
Example
input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i,j)=(2,3) twice and after that choose (i,j)=(1,2) twice too. These operations will transform [1,−1,0]→[1,−1,−2]→[1,1,−2]
In the second test case we can’t make equal numbers on the second position.
In the third test case we can choose (i,j)=(1,2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
思路
记录第一个1和-1的位置，根据两数组对应数字比较大小，两数组数字对比不同时比较所记录的1或-1位置。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll a[N], b[N];
int main()
{
int t, n;
cin >> t;
while (t--)
{
cin >> n;
for (int i = 0; i < n; i++)
cin >> a[i];
for (int i = 0; i < n; i++)
cin >> b[i];
int cnt = 0, k1 = n, k2 = n;
for (int i = 0; i < n; i++)
{
if (a[i] == 1) k1 = min(k1, i);
if (a[i] == -1) k2 = min(k2, i);
if (a[i] == b[i]) cnt++;
if (a[i] < b[i] && k1 < i) cnt++;
if (a[i] > b[i] && k2 < i) cnt++;
}
if (cnt == n)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
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