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;
#define int long long

const int N = 301;
const int MOD = 998244353;

signed main(){

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

  static int dp[N][N*N + N];
  //i個目まで見て、Rで塗った辺の長さがjとなる場合の数
  //一辺の値が(R+G+B)/2以上ならば三角形は作れない
  static int dp2[N][N*N + N];
  //2色に塗り分けたとき、i個目まで見て、Rで塗った辺の長さがj
  //R=G=(R+G+B)/2,B=0の重複カウントを取り除く
  dp[0][0] = 1;
  dp2[0][0] = 1;
  for(int i = 0; i < n; i++){
    for(int j = 0; j < N*N; j++){
      (dp[i+1][j + a[i]] += dp[i][j]) % MOD;
      (dp[i+1][j] += dp[i][j] * 2) % MOD;
      (dp2[i+1][j + a[i]] += dp2[i][j]) % MOD;
      (dp2[i+1][j] += dp2[i][j]) % MOD;
    }
  }

  int ans = 1;
  for(int i = 0; i < n; i++){
    (ans *= 3) % MOD;
  }

  int ng = 0;
  for(int i = (sum+1)/2; i < N*N; i++){
    (ng += dp[n][i]) % MOD;
  }
  if(sum%2 == 0){
    ng = (ng - dp2[n][sum/2]) % MOD;
  }
  (ng *= 3) % MOD;

  ans = (ans - ng + MOD) % MOD;
  cout << ans << endl;

  return 0;
}