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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)


def dp_table(*counts):  # dp[x][y]を作りたかったら dp_table(xのサイズ,yのサイズ)
    if counts:
        return [dp_table(*counts[1:]) for _ in range(counts[0])]
    else:
        return 0


N = int(input())
A = [int(input()) for i in range(N)]
mod = 998244353
ans = pow(3, N, mod)
sum_A = sum(A)
dp = dp_table(N, sum_A + 1)
# dp[i][j]: i番目までの中から辺をいくつか選んで、長さがjになるパターン数

# init
dp[0][A[0]] = 1
dp[0][0] = 2  # 他の色のうちどちらに入れるかで2通り

# init2
dp2 = dp_table(N, sum_A+1)  # ある二色でsum_Aを等分するパターンを数える
dp2[0][A[0]] = 1
dp2[0][0] = 1  # 今度は片方に集中したパターン

# dp
sum_now = A[0]
for i in range(1, N):
    a = A[i]
    sum_now += a
    for length in range(0, sum_now+1):
        # A[i]を追加しないのなら他の２食のうちどちらに入れるかで2倍選択肢がある
        if length >= a:
            dp[i][length] = dp[i - 1][length - a] + dp[i - 1][length] * 2
        else:
            dp[i][length] = dp[i-1][length]*2
        while (dp[i][length] > mod):
            dp[i][length] -= mod
        #dp[i][length] %= mod

    for length in range(min(sum_now + 1, sum_A // 2 + 1)):
        if length >= a:
            dp2[i][length] = dp2[i - 1][length - a] + dp2[i - 1][length]
        else:
            dp2[i][length] = dp2[i-1][length]
        while (dp2[i][length] > mod):
            dp2[i][length] -= mod
        #dp2[i][length] %= mod
# print(sum(dp[N-1][sum_A//2]))
# print(dp)
ans -= 3*sum(dp[N-1][(sum_A+1)//2:])
if sum_A % 2 == 0:
    ans += 3*dp2[N-1][sum_A//2]
print(ans % mod)