cati rime first print examples emp code long 中間

E. Magic Stones time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

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

ii, where 2≤i≤n−12≤i≤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 cici changes to c′i=ci+1+ci−1−cici′=ci+1+ci−1−ci.

Andrew, Grigory‘s friend, also has nn stones with charges titi. 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 cici into titi for all ii?

Input

The first line contains one integer

nn (2≤n≤1052≤n≤105) — the number of magic stones.

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.

Output

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

Otherwise, print "No".

Examples input Copy

4
7 2 4 12
7 15 10 12

output Copy

Yes

input Copy

3
4 4 4
1 2 3

output Copy

No

Note

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：

通過樣例觀察：

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

我們來看最初的數組，和中途的數組，以及目標數組，他們的差分都是【5，8，2】這三個數，變來變去都是這三個，

再加以觀察可以發現，我們每執行一個操作，影響的只是交換了差分，那麽只需要數組的首尾兩個數相等，並且中間的差分數排序後相等即可保證一一定能交換成功。

細節見代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rt return
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ‘\0‘, sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int a[maxn];
int b[maxn];
int main()
{
    gbtb;
    cin>>n;
    repd(i,1,n)
    {
        cin>>a[i];
    }
    repd(i,1,n)
    {
        cin>>b[i];
    }
    
    
    std::vector<int> v1;
    std::vector<int> v2;
    bool isok=1;
    if(a[1]!=b[1]||a[n]!=b[n])
    {
        // db(2);
        isok=0;
    }
    
    repd(i,2,n)
    {
        v1.pb(a[i]-a[i-1]);
        v2.pb(b[i]-b[i-1]);
    }
    int z=sz(v1);
    sort(v1.begin(),v1.end());
    sort(v2.begin(),v2.end());
    repd(i,0,z-1)
    {
        if(v1[i]!=v2[i])
        {
            isok=0;
        }
    }
    if(isok)
    {
        printf("Yes\n");
    }else
    {
        printf("No\n");
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ‘ ‘ || ch == ‘\n‘);
    if (ch == ‘-‘) {
        *p = -(getchar() - ‘0‘);
        while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
            *p = *p * 10 - ch + ‘0‘;
        }
    }
    else {
        *p = ch - ‘0‘;
        while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
            *p = *p * 10 + ch - ‘0‘;
        }
    }
}

Magic Stones CodeForces - 1110E （思維+差分）