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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const long long MOD = 998244353;

void solve(long long N, std::vector<long long> a){
    int dp[301][90301] = {};
    int dp2[301][90301] = {};
    int sum_a = std::accumulate(a.begin(), a.end(), 0);

    dp[0][0] = 1;
    dp2[0][0] = 1;

    for(int i = 0; i < N; i++)
    {
        for(long j = 0; j <= sum_a; j++)
        {
            dp[i+1][j] += dp[i][j] * 2;
            dp[i+1][j] %= MOD;
            dp[i+1][j+a[i]] += dp[i][j];
            dp[i+1][j+a[i]] %= MOD;

            dp2[i+1][j] += dp2[i][j];
            dp2[i+1][j] %= MOD;
            dp2[i+1][j+a[i]] += dp2[i][j];
            dp2[i+1][j+a[i]] %= MOD;
        }
        
    }
    
    long long border = (sum_a + 1) / 2;
    long long sum_r = 0;
    for(long j = border; j <= sum_a; j++)
    {
        sum_r += dp[N][j];
    }    
    
    long long result;
    result = 1;
    for(int i = 0; i < N; i++)
    {
        result = (result * 3) % MOD;
    }
    
    result = (result - 3 * sum_r) % MOD;
    if (border * 2 == sum_a) {
        result += 3 * dp2[N][border];
        result %= MOD;
    }
    if (result < 0) {
        result += MOD;
    }
    

    cout << result << endl;
}

// Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
int main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> a(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&a[i]);
    }
    solve(N, std::move(a));
    return 0;
}