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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ld long double
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
#define eb emplace_back
template<class T>void chmax(T &a,T b){if(a<b)a=b;}
template<class T>void chmin(T &a,T b){if(a>b)a=b;}
constexpr int INF=1000000000000000000;
constexpr int mod=998244353;
constexpr double eps=1e-8;
const double pi=acos(-1);
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int Random(int mi,int ma){
    random_device rnd;
    mt19937 mt(rnd());//32bit
    //[mi,ma]
    uniform_int_distribution<int>engine(mi,ma);
    return engine(mt);
}
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
vector<int>kaijo,invkaijo;
void init_fact(int n){
    kaijo.resize(n+1);
    invkaijo.resize(n+1);
    kaijo[0]=1;
    for(int i=1;i<=n;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
    rep(i,n+1)invkaijo[i]=modpow(kaijo[i],mod-2);
}
int comb(int a,int b){
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return kaijo[a]*modpow(kaijo[a-b],mod-2)%mod*modpow(kaijo[b],mod-2)%mod;
}
int inv(int x){
    x=modpow(x,mod-2);
    return x;
}
int n,a[310],ans=1,sum,dp[310][90010],dp2[310][90010];
signed main(){
    cin.tie(0);ios::sync_with_stdio(false);
    cin>>n;
    rep(i,n){
        cin>>a[i];
        sum+=a[i];
        ans*=3;
        ans%=mod;
    }
    dp[0][0]=3,dp2[0][0]=3;
    rep(i,n){
        rep(j,90000){
            if(j+a[i]<=90000){
                dp[i+1][j+a[i]]+=dp[i][j];
                dp[i+1][j+a[i]]%=mod;
                dp2[i+1][j+a[i]]+=dp2[i][j];
                dp2[i+1][j+a[i]]%=mod;
            }
            dp[i+1][j]+=dp[i][j]*2%mod;
            dp[i+1][j]%=mod;
            dp2[i+1][j]+=dp2[i][j];
            dp2[i+1][j]%=mod;
        }
    }
    for(int i=(sum+1)/2;i<=sum;i++){
        ans-=dp[n][i]%mod;
        ans+=mod;
        ans%=mod;
    }
    if(sum%2==0)ans+=dp2[n][sum/2]%mod;
    ans%=mod;
    cout<<ans<<endl;
    return 0;
}