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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

'use strict'
function Main(input){
  const K=998244353;
  input=input.trim().split(/\s+/).map(x=>x-0);
  var N=input.shift();
  var sum=input.reduce((a,c)=>a+c);
  var half=Math.floor(sum/2);
  var factor=[3];
  for(let i=0;i<N-1;i++){
    factor.push(factor[i]*2%K);
  }
  
  var dp=[];
  //dp[i][j][k]でinput[0~i]のうちj個を使って合計をkにする場合の数
  for(let i=0;i<=N;i++){
    dp[i]=[];
    for(let j=0;j<=i;j++){
      dp[i][j]=[];
      for(let k=0;k<=sum;k++){
        dp[i][j][k]=0;
      }
    }
  }
  dp[0][0][0]=1;
  for(let i=0;i<N;i++){
    for(let j=0;j<=i;j++){
      for(let k=0;k<=sum;k++){
        dp[i+1][j][k]+=dp[i][j][k];
        dp[i+1][j][k]%=K;
      }
      for(let k=input[i];k<=sum;k++){
        dp[i+1][j+1][k]+=dp[i][j][k-input[i]];
        dp[i+1][j+1][k]%=K;
      }
    }
  }
  var ans=1;
  for(let i=0;i<N;i++){
    ans*=3;
    ans%=K;
  }
  for(let i=0;i<N;i++){
    ans-=factor[i]*dp[N][i].reduce(function(b,c,j){
      if(j<=half)
        b+=c;
        b%=K;
      return b;
    },0)%K;
  }
  while(ans<0)
    ans+=K;
  ans%=K;
  if(sum%2===0){
    for(let j=0;j<=N;j++){
      ans+=3*dp[N][j][half];
      ans%=K;
    }
  }
  console.log(ans);
}

Main(require('fs').readFileSync('/dev/stdin','utf8'));