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 <iostream>
#include <algorithm>
#include <vector> //動的配列
#include <string>
#include <bits/stdc++.h>

using namespace std;

using ll=long long;

template <class T>ostream &operator<<(ostream &o,const vector<T>&v)
{o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}

//余事象で答えを求める

int n;
const ll mod=998244353;
ll l[330];
ll dp1[330][100001];
ll dp2[330][100001];

int main(){
    cin>>n;
    ll l[330];
    ll s=0;
    ll ans=1;
    const ll mod=998244353;
    for (int i=0;i<n;i++){
        cin>>l[i];
        s+=l[i];
        ans*=3;
        ans%=mod;
    }
    dp1 [0][0]=1;
    dp2 [0][0]=1;
    for (int i=0;i<n;i++){
        for (int j=0;j<=s;j++){
            dp1[i+1][j+l[i]] += dp1[i][j]; // 色を塗り替えても同じ．
            dp1[i+1][j+l[i]] %= mod;
            dp1[i+1][j] += (2*dp1[i][j])%mod;
            dp1[i+1][j] %= mod;

            dp2[i+1][j+l[i]] += dp2[i][j];
            dp2[i+1][j+l[i]] %= mod;

            dp2[i+1][j] += dp2[i][j];
            dp2[i+1][j] %= mod;
        }
    }
    for (int i=(s+1)/2;i<=s;i++){
        ans-=(3*dp1[n][i])%mod;
    }
    if (s%2==0){
        ans+=3*dp2[n][s/2];
    }
    ans+=mod;
    ans%=mod;
    cout<<ans<<endl;
    
}