题解

思考了一下,只要枚举前缀是否为总和的1/3即可

void solve() {
    int n;
    cin >> n;
    vl a(n + 1);
    vl s(n + 1, 0);

    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        s[i] = s[i - 1] + a[i];
    }
    if (n < 3 || s[n] % 3 != 0) {
        cout << 0 << "\n";
        return;
    }

    ll cnt1 = s[n] / 3;
    ll cnt2 = 2 * cnt1;
    ll ans = 0;
    ll cnt = 0;

    for (int i = 1; i < n; i++) {
        if (s[i] == cnt2) {
            ans += cnt;
        }
        if (s[i] == cnt1) {
            cnt++;
        }
    }
    cout << ans << "\n";
}