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

ACM_e發表於2017-12-14


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
4
6 6 5 5
output
0
input
4
6 6 4 4
output
-8
Note

In the first example:

  1. d(a1, a2) = 0;
  2. d(a1, a3) = 2;
  3. d(a1, a4) = 0;
  4. d(a1, a5) = 2;
  5. d(a2, a3) = 0;
  6. d(a2, a4) = 0;
  7. d(a2, a5) = 0;
  8. d(a3, a4) =  - 2;
  9. d(a3, a5) = 0;
  10. d(a4, a5) = 2.



此題爆了long long（9*1e18）,可以考慮用long double儲存，輸出用%Lf，但需要在C++14下才可以

考慮每個數對答案的貢獻，即被當做x和y的次數，然後再處理一下特殊的情況即可


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  爆這組











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