Codeforces 612B HDD is Outdated Technology

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B. HDD is Outdated Technology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 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 test(s)
input 
3
3 1 2

output 
3

input 
5
1 3 5 4 2

output 
10

Note
In the second example the head moves in the following way:
1->2 means movement from the sector 1 to the sector 5, i.e. it takes 4 time units
2->3 means movement from the sector 5 to the sector 2, i.e. it takes 3 time units
3->4 means movement from the sector 2 to the sector 4, i.e. it takes 2 time units
4->5 means movement from the sector 4 to the sector 3, i.e. it takes 1 time units
So the answer to the second example is 4 + 3 + 2 + 1 = 10.
제목: 1 - n의 전체 배열을 정하고 1 - 2 - 3 -... -n이동이 몇 번 필요합니까?
sort 한 번, 직접 통계하면 돼요.
AC 코드:
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (200000+10)
#define MAXM (200000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d
", (a))
#define Pf(a) printf("%.2lf
", (a))
#define Pl(a) printf("%lld
", (a))
#define Ps(a) printf("%s
", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#define first fi
#define second se
using namespace std;
struct Node{
    int val, id;
};
Node num[MAXN];
bool cmp(Node a, Node b){
    return a.val < b.val;
}
int main()
{
    int n; Ri(n);
    for(int i = 0; i < n; i++)
    {
        Ri(num[i].val);
        num[i].id = i;
    }
    sort(num, num+n, cmp);
    LL ans = 0;
    for(int i = 0; i < n-1; i++)
        ans += abs(num[i].id - num[i+1].id);
    Pl(ans);
    return 0;
}
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