传送门

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将原数组差分一下，设\(d_i = c_{i+1} - c_i\)

考虑在\(i\)位置的一次操作会如何影响差分数组

\(d_{i+1}' = c_{i+1} - (c_{i+1} + c_{i-1} - c_i) = c_i - c_{i-1} = d_i\)

\(d_i' = (c_{i+1} + c_{i-1} - c_i) - c_{i-1} = c_{i+1} - c_i = d_{i+1}\)

所以本质上一次操作交换了差分数组中的两个元素

所以只需要判断差分数组中的每一个元素能否对应即可。注意还需要判断\(c_1=t_1\)

#include<bits/stdc++.h>

using namespace std;

inline int read(){

    int a = 0;

    char c = getchar();

    bool f = 0;

    while(!isdigit(c)){

        if(c == '-')

            f = 1;

        c = getchar();

    }

    while(isdigit(c)){

        a = (a << 3) + (a << 1) + (c ^ '0');

        c = getchar();

    }

    return f ? -a : a;

}

const int MAXN = 1e5 + 7;

int c[MAXN] , t[MAXN] , N; 

int main(){

    N = read();

    for(int i = 1 ; i <= N ; ++i)

        c[i] = read();

    for(int i = 1 ; i <= N ; ++i)

        t[i] = read();

    if(c[1] != t[1]){

        puts("No");

        return 0;

    }

    for(int i = 1 ; i < N ; ++i){

        c[i] = c[i + 1] - c[i];

        t[i] = t[i + 1] - t[i];

    }

    sort(c + 1 , c + N);

    sort(t + 1 , t + N);

    for(int i = 1 ; i < N ; ++i)

        if(c[i] != t[i]){

            puts("No");

            return 0;

        }

    puts("Yes");

    return 0;

}