【差分】Day6-D

链接

D

题目描述

多个询问，每次询问给出长度为n的序列a,b,每次可以让ai’ = a[i+1] + a[i-1]-a[i]，问a能否变为b

思路

$a_i^{\prime }-a_{i-1}=a_{i+1}-a_i$
$a_i-a_{i-1}=a_{i+1}-a_i^{\prime }$
那么就是a数组的左右差分交换
比较a数组和b数组的差分就好了

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

int T, n, a[100005], b[100005], d[100005], d2[100005];

int read() 
{
    
    
	int x = 0, flag = 1; char ch = getchar();
	while (ch < '0' || ch > '9') {
    
    if (ch == '-') flag = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9'){
    
    x = x * 10 + ch - '0'; ch = getchar();}
	return x * flag;
}

int main()
{
    
    
	T = read(); 
	for(int i = 1; i <= T; ++i)
	{
    
    
		bool c = 0;
		n = read();
		for(int j = 1; j <= n; ++j)
			a[j] = read(), d[j] = a[j] - a[j - 1];
		for(int j = 1; j <= n; ++j)
			b[j] = read(), d2[j] = b[j] - b[j - 1];
		sort(d + 1, d + n + 1);
		sort(d2 + 1, d2 + n + 1);
		for(int j = 1; j <= n; ++j)
			if(d[j] != d2[j]) {
    
    
				printf("No\n");
				c = 1;
				break;
			}
		if(!c) printf("Yes\n");
		
	}
}