CF1495E Qingshan and Daniel

场上不到半小时就过了，感觉这题的难度取决于它是第几题。。

首先注意到加起来小的阵营会把牌打完，现在考虑小阵营对大阵营的贡献。

每个小阵营机器人必然会以顺时针的方向对后面的大阵营机器人做贡献，那么我们直接绕着环跑两圈，记录当前小阵营打出的牌数和往后做贡献就行了，至于为啥要跑两圈是因为第一圈可能没贡献干净。

时间复杂度 $\mathcal{O}(n)$。

代码：

int main () {
  // freopen ("1.in", "r", stdin);
  // freopen ("1.out", "w", stdout);
  n = rd (), m = rd ();
  rep (i, 1, m) p[i] = rd (), k[i] = rd (), b[i] = rd (), w[i] = rd ();
  generate ();
  ll s[2] = {0, 0};
  rep (i, 1, n) s[t[i]] += a[i];
  ll now (0);
  bool o (s[0] > s[1]);
  if (t[1] != o) ++ ans[1], -- a[1];
  rep (i, 1, n) {
    if (t[i] == o) now += a[i], ans[i] += a[i], a[i] = 0;
    else {
      if (now >= a[i]) ans[i] += a[i], now -= a[i], a[i] = 0;
      else ans[i] += now, a[i] -= now, now = 0;
    }
  }
  rep (i, 1, n) {
    if (t[i] == o) now += a[i], ans[i] += a[i], a[i] = 0;
    else {
      if (now >= a[i]) ans[i] += a[i], now -= a[i], a[i] = 0;
      else ans[i] += now, a[i] -= now, now = 0;
    }
  }
  ll ret (1);
  rep (i, 1, n) ret = (ret * (((1ll * i * i) ^ ans[i]) % P + 1)) % P;
  cout << ret;
}