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>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=m-1; i>=n; i--)
#define SORT(v,n) do{sort(v,v+n);reverse(v,v+n);}while(0)
#define all(x) (x).begin(),(x).end()
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 998244353;
typedef pair<ll, ll> LP;

ll POW(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW(x*x,n/2)%MOD;
  return x%MOD*POW(x,n-1)%MOD;
}

ll POW2(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW2(x*x,n/2);
  return x*POW2(x,n-1);
}

ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll KAI2(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI2(m-1);
}

ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI(m)%MOD*modinv(KAI(n)%MOD*KAI(m-n)%MOD)%MOD;
}

ll COM2(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI2(m)/KAI2(n)/KAI2(m-n);
}

ll DEC(ll x,ll m,ll n)
{
  return x%POW(m,n+1)/POW(m,n);
}

ll keta(ll x,ll n)
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}

ll DIV(ll x,ll n)
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}

ll ORD(ll x,ll n)
{
  if(x==0)return INF;
  if(x%n!=0)return 0;
  return 1+ORD(x/n,n);
}

int main()
{
  ll n,a[500],dp[3][120000]={},dp2[3][120000]={},s=0,t,ans=0;
  dp[0][0]=1;
  dp2[0][0]=1;
  cin >> n;
  rep(i,n)cin >> a[i];
  rep(i,n)s+=a[i];
  t=(s+1)/2;
  rep(i,n){
    rep(j,100000){
      if(j-a[i]>=0)dp[1][j]=(dp[0][j]*2+dp[0][j-a[i]])%MOD;
      else dp[1][j]=dp[0][j]*2%MOD;
    }
    rep(j,100000)dp[0][j]=dp[1][j];
  }
  rep(i,n){
    rep(j,100000){
      if(j-a[i]>=0)dp2[1][j]=(dp2[0][j]+dp2[0][j-a[i]])%MOD;
      else dp2[1][j]=dp2[0][j]%MOD;
    }
    rep(j,100000)dp2[0][j]=dp2[1][j];
  }
  FOR(i,t,100000)ans=(ans+dp[1][i])%MOD;
  if(s%2==0)ans=((ans-dp2[1][s/2]*2)%MOD*3+dp2[1][s/2]*3)%MOD;
  else ans=ans*3%MOD;
  ans=((POW(3,n)-ans)%MOD+MOD)%MOD;
  printf("%lld",ans);
}