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

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本文介绍了一种针对数组中所有元素对计算AlmostDifference并求和的算法实现。该问题要求计算所有可能的元素对(i, j) (1 ≤ i ≤ j ≤ n)的d(ai, aj)之和。通过分析每个元素对答案的贡献度，文章提供了一个有效的解决方案，并附带了C++代码示例。

Let's denote a function

$\text{[math]}$

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
4
6 6 5 5

output
0

input
4
6 6 4 4

output
-8

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.

此题爆了long long（9*1e18）,可以考虑用long double存储，输出用%Lf，但需要在C++14下才可以

考虑每个数对答案的贡献，即被当做x和y的次数，然后再处理一下特殊的情况即可

蠢哭了

看了一天

#include<bits/stdc++.h>
using namespace std;
#define maxn 200000+1000
#define LL long long
long long a[maxn];
map<long long,int>q;
int main(){
   int n;
   cin>>n;
   long double sum=0;
   for(int j=1;j<=n;j++){
      cin>>a[j];
      sum+=(j-1)*a[j];
      sum-=(n-j)*a[j];
   }
   for(int j=1;j<=n;j++){
      q[a[j]]++;
      sum-=q[a[j]-1];
      sum+=q[a[j]+1];
   }
   printf("%.0Lf\n",sum);
}
//-9999999990000000000   longlong  爆这组