Problem:
Score : 600 points

Problem Statement

You are given N integers. The i-th integer is a_i.
Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:

Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.

Constraints

3 \leq N \leq 300

1 \leq a_i \leq 300(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
:
a_N

Output

Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.

Sample Input 1

4
1
1
1
2

Sample Output 1

18

We can only paint the integers so that the lengths of the sides of the triangle will be 1, 2 and 2, and there are 18 such ways.

Sample Input 2

6
1
3
2
3
5
2

Sample Output 2

150

Sample Input 3

20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4

Sample Output 3

563038556

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll M = 998244353;

int main() {
    int N;
    cin >> N;
    vector<ll> a(N);
    ll sum = 0;
    for (int i = 0; i < N; i++) {
        cin >> a[i];
        sum += a[i];
    }

    vector<vector<ll>> dp(N + 1, vector<ll>(sum + 301, 0));
    dp[0][0] = 1;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= sum; j++) {
            dp[i + 1][j + a[i]] += dp[i][j];
            dp[i + 1][j + a[i]] %= M;
            dp[i + 1][j] += 2 * dp[i][j];
            dp[i + 1][j] %= M;
        }
    }

    vector<vector<ll>> dp2(N + 1, vector<ll>(sum, 0));
    dp2[0][0] = 1;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j <= sum / 2; j++) {
            dp2[i + 1][j + a[i]] += dp2[i][j];
            dp2[i + 1][j + a[i]] %= M;
            dp2[i + 1][j] += dp2[i][j];
            dp2[i + 1][j] %= M;
        }
    }

    ll x = 0;
    for (int i = (sum + 2 - 1) / 2; i <= sum; i++) {
        x += dp[N][i];
        x %= M;
    }

    ll ans = pow(3, N);
    ans -= 3 * x;
    if (sum % 2 == 0) ans += 3 * dp2[N][sum / 2];

    cout << ans % M << endl;
}