【Codeforces Global Round 1 E】Magic Stones

【链接】 我是链接,点我呀:)
【题意】

你可以把c[i]改成c[i+1]+c[i-1]-c[i] (2<=i<=n-1) 问你能不能把每一个c[i]都换成对应的t[i];

【题解】

```cpp d[i] = c[i+1]-c[i]; (1<=i<=n-1)

		change c[i]
		
		c[i]' = c[i+1]+c[i-1]-c[i];
		
		d[i-1] = c[i]'-c[i-1];
		       = c[i+1]+c[i-1]-c[i]-c[i-1] == c[i+1]-c[i] = d[i];
		
		d[i] = c[i+1]-c[i]'
			 = c[i+1]-c[i+1]-c[i-1]+c[i]
			 = c[i] - c[i-1];
			 = d[i-1];
也就是说对2..n-1进行操作的话
就是把相邻的d的值交换一下
    显然这样的交换能让d变成任意顺序
因此,只要c[1]==d[1]并且c[n]==d[n]
然后排序后的c和排序后的d相同的话,就ok.

这里必须要用Integer的equals方法比较才不会超时。。
直接用int的a[i]!=b[i]会超时>_<

</font>


<font color = black size = 6> 【代码】</font>
```cpp
import java.io.*;
import java.util.*;

//code start from here
/*
	d[i] = c[i+1]-c[i]; (1<=i<=n-1)
	
	change c[i]
	
	c[i]' = c[i+1]+c[i-1]-c[i];
	
	d[i-1] = c[i]'-c[i-1];
	       = c[i+1]+c[i-1]-c[i]-c[i-1] == c[i+1]-c[i] = d[i];
	
	d[i] = c[i+1]-c[i]'
		 = c[i+1]-c[i+1]-c[i-1]+c[i]
		 = c[i] - c[i-1];
		 = d[i-1];
 */

public class Main {
	
	final static int N = (int)1e5;
	static InputReader in;
	static PrintWriter out;
	static int n;
	static Integer c[],t[],d1[],d2[];
		
	public static void main(String[] args) throws IOException{
		in = new InputReader();
		//out = new PrintWriter(System.out);
		c = new Integer[N+10];t = new Integer[N+10];
		d1 = new Integer[N+10]; d2 = new Integer[N+10];
		

		n = in.nextInt();
		for (int i = 1;i <= n;i++) c[i] = in.nextInt();
		for (int i = 1;i <= n;i++) t[i] = in.nextInt();
		if (!c[1].equals(t[1]) || !c[n].equals(t[n])){
			System.out.println("No");
			return;
		}
		for (int i = 1;i <= n-1;i++) d1[i] = c[i+1]-c[i];
		for (int i = 1;i <= n-1;i++) d2[i] = t[i+1]-t[i];
		Arrays.sort(d1, 1,n);
		Arrays.sort(d2, 1,n);
		for (int i = 1;i <= n-1;i++)
			if (!d1[i].equals(d2[i])) {
				System.out.println("No");
				return;
			}
		System.out.println("Yes");
	}

	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		
		public InputReader() {
			br = new BufferedReader(new InputStreamReader(System.in),32768);
			tokenizer = null;
		}
		
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
				tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}