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.
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.
Print one integer — the sum of d(ai, aj) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.
51 2 3 1 3
4
46 6 5 5
0
46 6 4 4
-8
In the first example:
【题意】:计算每个数贡献的值之和。贡献规则如上分段函数所示。
【分析】:假设第i位值是a,那么他的贡献tmp=a*(i-1)-presum。tmp是假设都计算为y-x,presum是前缀和。那最后的贡献就是tmp-cnt(a-1)+cnt(a+1),相邻的数去掉。因为相邻的数和本身总是相差1,符合分段函数归0部分。前面出现过多少次a-1,a+1,用map实现。
【代码】:
#include <bits/stdc++.h> using namespace std; map<double,double>a; int main() { int n; scanf("%d",&n); long double ans=0; for (int i=0;i<n;i++) { double x;scanf("%lf",&x); a[x]++; ans= ans+ a[x+1]-a[x-1]+x*(i+1-n+i); } cout << fixed << setprecision(0) << ans << endl; return 0; }