思路：dp

设sum为所有边的总和

不能组成三角形的情况：某条边长度>=ceil(sum/2)，可以用dp求出这种情况的方案数，然后用总方案数减去就可以求出答案。

注意当某两条边都为sum/2的时候，dp会多算一次，要减去多算的方案数，多算的方案数也可以用dp求

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
     
      using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r//#define mp make_pair#define pb push_back#define ULL unsigned LL#define pll pair
      
       #define pli pair
       
        #define pii pair
        
         #define piii pair
         
          #define pdi pair
          
           #define pdd pair
           
            #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int N = 305, M = 9e4 + 10;const int MOD = 998244353;int a[N], dp[N][M], pp[M], n, s = 0;LL q_pow(LL n, LL k) { LL res = 1; while(k) { if(k&1) res = (res * n) % MOD; n = (n * n) % MOD; k >>= 1; } return res;}int main() { scanf("%d", &n); for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), s += a[i]; dp[0][0] = 1; for (int i = 1; i <= n; ++i) { for (int j = 0; j < M; ++j) { dp[i][j] = (2*dp[i-1][j]) % MOD; } for (int j = a[i]; j < M; ++j) { dp[i][j] = (dp[i][j] + dp[i-1][j-a[i]]) % MOD; } } pp[0] = 1; for (int i = 1; i <= n; ++i) { for (int j = M-1; j >= a[i]; --j) pp[j] = (pp[j] + pp[j-a[i]]) % MOD; } if(s%2 == 0)dp[n][s/2] = (dp[n][s/2]-pp[s/2]) % MOD; LL ans = q_pow(3, n); int up = (s+1)/2; for (int i = up; i < M; ++i) ans = (ans - dp[n][i]*3LL%MOD) % MOD; printf("%lld\n", (ans + MOD) % MOD); return 0;}