【ACWing】913. Queue to fetch water

Subject address:

https://www.acwing.com/problem/content/description/915/

Have $n$ people line up to fetch water,$i\; It$ takes time for an individual to fetch water$t_i$, Ask how to arrange the order of fetching water to minimize the total waiting time for everyone.

Input format: the
first line contains the integer $n$ . The second line contains$n$ integers, where the$i$ integers represent the$i$ the time it takes for an individual to fill the bucket$t_i$。

Output format:
output an integer, representing the sum of the minimum waiting time.

Data range:
$1\le n\le 10^5$
$1\le t_i\le 10^4$

First to $t_i$Sorting from small to large, the total waiting time for fetching water in this order is the least.

Proof of algorithm correctness:
method of proof by contradiction. If the order of arrangement is not $t_i$Monotonically increasing, then there must be a certain $k$ such that$t_k>t_{k+1}$, Then swap the order in which these two people fetch water, the total fetch time will be reduced by $t_k-t_{k+1}>0$ , contradiction.

code show as below:

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100010;
int a[N];

int main() {
    
    
    int n;
    cin >> n;

    for (int i = 0; i < n; i++) cin >> a[i];

    sort(a, a + n);

    long res = 0;
    for (int i = 0; i < n; i++) res += (long) (n - i - 1) * a[i];

    cout << res << endl;

    return 0;
}

Time complexity $O\left(n\log n\right)$ , space$O\left(1\right)$。