CodeForces612BHDD is Outdated Technology（sort，结构体，水题）

Description

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

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

Find the time need to read file split to n fragments. The i-th sector contains the fi-th fragment of the file (1 ≤ fi ≤ n). Note different sectors contains the different fragments. At the start the magnetic head is in the position that contains the first fragment.
The file are reading in the following manner: at first the first fragment is read, then the magnetic head moves to the sector that contains the second fragment, then the second fragment is read and so on until the n-th fragment is read. The fragments are read
in the order from the first to the n-th.

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

Input

The first line contains a positive integer n (1 ≤ n ≤ 2·105) — the number of fragments.

The second line contains n different integers fi (1 ≤ fi ≤ n) — the number of the fragment written in the i-th sector.

Output

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

Sample Input

Input

3

3 1 2

Output

3

Input

5

1 3 5 4 2

Output
10

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
struct st{
__int64 nm;
__int64 q;
}a[200001];
bool cmp(st x,st y)
{
return x.nm<y.nm;
}
int main()
{
__int64 n;
while(scanf("%I64d",&n)!=EOF)
{
for(__int64 i=0;i<n;i++)
{
scanf("%I64d",&a[i].nm);
a[i].q=i;
}
sort(a,a+n,cmp);
__int64 sum=0;
for(__int64 i=0;i<n;i++)
{
if(i!=n-1)
sum+=abs(a[i+1].q-a[i].q);
}
printf("%I64d\n",sum);
}
return 0;
}思路：
给你一个打乱顺序的数组n，让你从数值为1的数一直按顺序走到n，问总共走多少步。

直接将数保存在结构体数组内，并在结构体内另保存一个数记录初始地址。sort快排后利用保存的初始地址算步数，然后输出就可以了。