Educational Codeforces Round 34 (Rated for Div. 2)D. Almost Difference(数学)

D. Almost Difference

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Let's denote a function



You are given an array a consisting of n integers.
You have to calculate the sum of d(ai, aj) over
all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Input

The first line contains one integer n (1 ≤ n ≤ 200000)
— the number of elements in a.

The second line contains n integers a1, a2,
..., an (1 ≤ ai ≤ 109)
— elements of the array.

Output

Print one integer — the sum of d(ai, aj) over
all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Examples

input

5
1 2 3 1 3

output

4

input

46 6 5 5

output

0

input

46 6 4 4

output

-8

题目大意:有一个序列，现在根据要求求出最后d(x,y)总和

解题思路:考虑到这个题不能暴力，这个题要做的是求一个前缀和，同时判断在当前值ai之前有多少个比ai大1和比ai小1的个数，然后计算

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <iomanip>

using namespace std;
typedef long long LL;
typedef long double ld;
ld cnt,sum[200005];
map<LL,int> a;
LL x;
int i,n;
int main()
{
cin>>n;
cnt=0;
for(i=1;i<=n;i++)
{
cin>>x;
cnt+=x*(i-1)-sum[i-1];
if(a[x-1])
cnt-=a[x-1];
if(a[x+1])
cnt+=a[x+1];
sum[i]=sum[i-1]+x;
a[x]++;
}
cout << fixed << setprecision(0) << cnt << endl;
}

Note

In the first example:

d(a1, a2) = 0;

d(a1, a3) = 2;

d(a1, a4) = 0;

d(a1, a5) = 2;

d(a2, a3) = 0;

d(a2, a4) = 0;

d(a2, a5) = 0;

d(a3, a4) =  - 2;

d(a3, a5) = 0;d(a4, a5) = 2.