CodeForces 612B HDD is Outdated Technology（磁盘顺道寻址问题）

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本文介绍了一种计算HDD硬盘上文件碎片读取时间的方法。通过追踪磁头移动路径，计算不同碎片间的距离来得出总读取时间。文章包含了一个示例性的C++实现代码。

B. HDD is Outdated Technology

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

HDD hard drives groupdata by sectors. All files are split to fragments and each of them are writtenin some sector of hard drive. Note the fragments can be written in sectors inarbitrary order.

One of the problemsof HDD hard drives is the following: the magnetic head should move from onesector to another to read some file.

Find the time need toread file split ton fragments. Thei-thsector contains thef_i-thfragment of the file (1 ≤ f_i ≤ n).Note different sectors contains the different fragments. At the start themagnetic head is in the position that contains the first fragment. The file arereading in the following manner: at first the first fragment is read, then themagnetic head moves to the sector that contains the second fragment, then the secondfragment is read and so on until then-thfragment is read. The fragments are read in the order from the first to then-th.

It takes |a - b|time units to move the magnetic head from the sectorato the sectorb. Reading a fragment takes no time.

Input

The first linecontains a positive integern (1 ≤ n ≤ 2·10⁵)— the number of fragments.

The second linecontains ndifferent integersf_i(1 ≤ f_i ≤ n)— the number of the fragment written in the i-thsector.

Output

Print the only integer — the number oftime units needed to read the file.

Examples

Input

3
3 1 2

Output

Input

5
1 3 5 4 2

Output

Note

In the second examplethe head moves in the following way:

· 1->2means movement from the sector 1 to the sector 5, i.e. it takes 4 time units

· 2->3means movement from the sector 5 to the sector 2, i.e. it takes 3 time units

· 3->4means movement from the sector 2 to the sector 4, i.e. it takes 2 time units

· 4->5means movement from the sector 4 to the sector 3, i.e. it takes 1 time units

So the answer to the second example is4 + 3 + 2 + 1 = 10.

参考题意：

磁盘需要寻址检索，一个扇区接一个的顺序寻找，从 1 —2—3—4—n 需要的时间是两者之间的距离。

参考思路：

很水的一个题，用数组存储当前位置，数组下标为扇区号。

参考代码：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define MYDD 110300*2

using namespace std;

int main() {
	int n,k;
	int wqs[MYDD];
	__int64 ans;
	while(scanf("%d",&n)!=EOF) {
		for(int j=1; j<=n; j++) {
			scanf("%d",&k);
			wqs[k]=j;// k 磁道在 j 位置
		}

		ans=0;
		for(int j=1; j<n; j++) {
			ans+=abs(wqs[j]-wqs[j+1]);//循环遍历整个硬盘，但愿不超时
			//	printf("*****%d**%d**%d*\n",ans,j,j+1);
		}

		printf("%I64d\n",ans);
	}
	return 0;
}

/*
5
1 3 5 4 2
*/