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: Accepted)

# https://atcoder.jp/contests/tenka1-2019/submissions/5065329

import sys

input = sys.stdin.readline

N = int(input())
a = [int(input()) for _ in range(N)]
total = sum(a)
MOD = 998244353

'''(1)全体'''
U = pow(3, N, MOD)  # 三角形を構成できない場合を含むRGBの組み合わせ総数

'''(2)単円'''
dp = [[0] * (total + 1) for _ in range(N + 1)]
dp[0][0] = 1
# dp[i][R]:=i番目まででRとなる場合の数

for i in range(N):
    for r in range(total + 1):
        if r - a[i] >= 0:
            dp[i + 1][r] += dp[i][r - a[i]]
        dp[i + 1][r] += 2 * dp[i][r]
        dp[i + 1][r] %= MOD
        # g, bに加算する分で、2回起こる
# print(dp)

single_circle = 0
for r in range(total + 1):
    if r >= (total / 2):
        single_circle += dp[N][r]
# print(single_circle)

'''(3)２円の積集合'''
if total % 2 == 0:
    dp2 = [[0] * (total + 1) for _ in range(N + 1)]
    dp2[0][0] = 1
    # dp[i][R]:=i番目まででRとなる場合の数

    for i in range(N):
        for r in range(int(total / 2) + 1):
            if r - a[i] >= 0:
                dp2[i + 1][r] += dp2[i][r - a[i]]
            dp2[i + 1][r] += dp2[i][r]
            dp2[i + 1][r] %= MOD
    # print(dp2)

two_circle_intersect = dp2[N][int(total / 2)] if total % 2 == 0 else 0

'''(3)３円の積集合'''
three_circle_intersect = 0

'''(4)解答'''
ans = 0

ans += U
ans %= MOD

ans -= 3 * single_circle
ans %= MOD

ans += 3 * two_circle_intersect
ans %= MOD

ans -= three_circle_intersect
ans %= MOD

print(ans)