AtCoder Regular Contest 116 B

Problem Statement

Given is a sequence A of N integers. There are 2N−1 non-empty subsequences B of A. Find the sum of max(B)×min(B) over all of them.

Since the answer can be enormous, report it modulo 998244353.

Constraints

• All values in input are integers.
• 1≤N≤2×105
• 0≤Ai≤998244352

和2021牛客寒假训练赛某题很类似。暴力肯定不可做，考虑枚举贡献。首先对序列排序，朴素的想法是枚举最大值a[j]和最小值a[i],乘起来再乘以以此为最大值最小值的序列个数(即(2^{j - i - 1}))，注意如果j == i 或者j == i + 1的话个数只有一个。这样也是会t的。不妨进一步考虑，只枚举最大值，然后乘以一个类似前缀和的东西，比如枚举到的最大值是a[4]，那么此时答案要加上的就是(a[4] imes(a[1] imes2^2+a[2] imes2^1+a[3] imes2^0+a[4] imes2^0))，如果枚举到的是a[5]，则为(a[5] imes(a[1] imes2^3+a[2] imes2^2+a[3] imes2^1+a[4] imes2^0+a[5] imes2^0))。不难发现规律：不妨设i = 4时类前缀和tmp为(a[1] imes2^2+a[2] imes2^1+a[3] imes2^0)，当i = 5时tmp先变为2倍，然后再加上(a[i - 1] imes2^0)，此时令(ans += tmp + a[i] imes a[i])即可。注意特判i = 1，2，3的情况。

#include <iostream>
#define mod 998244353
using namespace std;
int n;
long long a[2000005], ans = 0;
int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	long long tmp = 0;
	for(int i = 1; i <= n; i++) {
		if(i == 1) {
			tmp = a[1];
			ans += a[1] * a[1] % mod;
		} else if(i == 2) {
			tmp = (a[1] + a[2]) % mod;
			ans = (ans + tmp * a[2] % mod) % mod;
		} else if(i == 3) {
			tmp = (2 * a[1] % mod + a[2] % mod) % mod;
			ans = (ans + tmp * a[3] % mod) % mod;
			ans = (ans + a[3] * a[3] % mod) % mod; 
		} else {
			tmp = tmp * 2 % mod;
			tmp = (tmp + a[i - 1]) % mod;
			ans = (ans + tmp * a[i] % mod) % mod;
			ans = (ans + a[i] * a[i] % mod) % mod;
		}
	}
	cout << ans;
	return 0;
}