Codeforces Round # 560 (Div. 3) E. Two arrays and the sum of functions (思维)

题目链接：http://codeforces.com/contest/1165/problem/E

题意：

改变数组B的顺序，使得最小

我们可以把 a[i] * (i + 1)*(n - i) 可以处理成定值,然后从大到小排序与从小到大排序的b数组相乘即为答案

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#define rep(i, s, e) for(int i = s; i < e; ++i)
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define P pair<int, int>
#define INF 0x3f3f3f3f
#define Mod 998244353
using namespace std;
typedef long long ll;
static const int N = 305;
static const int MAX_N = 2e5 + 5;
ll a[MAX_N], b[MAX_N];
void solve(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    //ios::sync_with_stdio(false);
    int n;
    while(scanf("%d", &n) != EOF){
        for(int i = 0; i < n; ++i){
            ll v;
            scanf("%I64d", &v);
            a[i] = v * (i + 1) * (n - i);    //这里不能先取模，不然后面排序的贪心策略就会有问题
        }
        for(int i = 0; i < n; ++i) scanf("%I64d", &b[i]);
        sort(a, a + n, greater<ll>());
        sort(b, b + n);
        ll ans = 0;
        for(int i = 0; i < n; ++i) ans = (ans + (a[i] % Mod) * b[i]) % Mod;
        printf("%I64d\n", ans);
    }
}
int main() {
    solve();
    return 0;
}