E. Two Arrays and Sum of Functions

 

题意：给定序列A，B。可以对B重排，使得 最小/

思路：可以求出每个a[i]的计算的次数然后加在a[i]上，然后排序计算。a[i]的计算次数是i*(n-i+1)次。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5 + 10;
const int mod = 998244353;
ll n, a[N], b[N], ans;
int cmp(ll a, ll b) {
    return a > b;
}
int main() {
    cin >> n;
    for(int i = 1; i <= n; i++)
        cin >> a[i], a[i] = a[i] * i * (n - i + 1);
    for(int i = 1; i <= n; i++)
        cin >> b[i];
    sort(b + 1, b + n + 1, cmp);
    sort(a + 1, a + n + 1);
    for(int i = 1; i <= n; i++) {
        ans = (ans + a[i] % mod * b[i] % mod) % mod;
    }
    cout << ans;
    return 0;
}