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：如果a[i]>b[i]的话，就看前面i有没有-1；如果a[i]<b[i]的话，就看前面i有没有1.类似于前缀那样。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=1e5+100;
int a[maxx];
int b[maxx];
int suma[maxx],sumb[maxx],sumc[maxx];
int n;
inline void init()
{
for(int i=0;i<=n;i++) suma[i]=sumb[i]=sumc[i]=0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
init();
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(a[i]==1) suma[i]=1,sumb[i]=sumb[i-1],sumc[i]=sumc[i-1];
if(a[i]==0) sumb[i]=1,suma[i]=suma[i-1],sumc[i]=sumc[i-1];
if(a[i]==-1) sumc[i]=1,suma[i]=suma[i-1],sumb[i]=sumb[i-1];
}
for(int i=1;i<=n;i++) cin>>b[i];
int flag=1;
for(int i=1;i<=n;i++)
{
if(a[i]>b[i]&&sumc[i-1]==0) flag=0;
else if(a[i]<b[i]&&suma[i-1]==0) flag=0;
if(flag==0) break;
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
努力加油a啊，(o)/~
