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

Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index

Andrew, Grigory‘s friend, also has n

The first line contains one integer

The second line contains integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤2?1090≤ci≤2?109 ) — the charges of Grigory‘s stones.

The second line contains integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤2?1090≤ti≤2?109 ) — the charges of Andrew‘s stones.

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

Otherwise, print "No".

4
7 2 4 12
7 15 10 12

Yes

3
4 4 4
1 2 3

No

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][7,2,4,12]→[7,2,10,12] .
• Then synchronize the second stone: [7,2,10,12]→[7,15,10,12][7,2,10,12]→[7,15,10,12] .

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

思路：

註意看題目給出的公式

c[i]‘=c[i+1]+c[i-1]-c[i]

可以變形成

c[i+1]-c[i]‘=c[i]-c[i-1]

c[i]‘-c[i-1]=c[i+1]-c[i]

這樣看來就是將第i項左右差分交換，可以把這個看成一種排序

所以將差分全部重新排序之後，如果上下差分都相同，則滿足題目要求

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
ll a[maxn],b[maxn],c[maxn],d[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
    for(int i=1;i<=n;i++) scanf("%I64d",&b[i]);
    for(int i=1;i<n;i++)
    {
        c[i]=a[i+1]-a[i];
        d[i]=b[i+1]-b[i];
    }
    sort(c+1,c+n);
    sort(d+1,d+n);
    if(a[1]!=b[1]||a[n]!=b[n])
    {
        printf("No\n");
        return 0;
    }
    for(int i=1;i<n;i++)
    {
        if(c[i]!=d[i])
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    return 0;
}

E. Magic Stones CF 思維題