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<bits/stdc++.h> 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<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<int, pii> #define pdi pair<double, int> #define pdd pair<double, double> #define mem(a, b) memset(a, b, sizeof(a)) #define debug(x) cerr << #x << " = " << x << " "; #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //head const 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 ", (ans + MOD) % MOD); return 0; }