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docs/ACM/Knowledge/Math/其他/xor.md

@@ -0,0 +1,125 @@
+## 技巧
+
+- 拆位
+- 公式
+
+## 牛客第七场 C Beautiful Sequence
+
+[题目链接](https://ac.nowcoder.com/acm/contest/57361/C)
+
+![](assets/image-20230807195320433.png)
+
+![](assets/2023-10-06-01-38-28.png)
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+void solve()
+{
+ int n, k;
+ cin >> n >> k;
+
+ vector<int> b(n + 1), w(n + 1);
+ for (int i = 1; i < n; i++)
+ {
+  cin >> b[i];
+  w[i + 1] = w[i] ^ b[i];
+ }
+
+ vector<int> a1(30, -1);
+ for (int i = 1; i < n; i++)
+ {
+  int u = -1;
+
+  for (int j = 29; j >= 0; j--)
+  if ((b[i] >> j) & 1)
+  {
+  u = j;
+  break;
+  }
+
+  // 当 Bi == 0 的时候直接跳过即可
+  if (u == -1)
+  continue;
+  // ==> a[i]的第u位为0
+  // ==> a[1]的第u位为t
+
+  int t1 = ((w[i] >> u) & 1) ^ 0;
+
+  if (a1[u] != -1 && a1[u] != t1)
+  {
+  cout << "-1" << endl;
+  return;
+  }
+  else
+  a1[u] = t1;
+ }
+
+ // for (int i = 29; i >= 0; i--)
+ // cerr << a1[i] << endl;
+
+ int cnt = 0;
+ for (int i = 0; i < a1.size(); i++)
+  if (a1[i] == -1)
+  cnt++;
+
+ if (pow(2, cnt) < k)
+ {
+  cout << "-1" << endl;
+  return;
+ }
+ k--;
+ while (k)
+ {
+  int c = k & 1;
+  k >>= 1;
+  for (int i = 0; i <= 29; i++)
+  if (a1[i] == -1)
+  {
+  a1[i] = c;
+  break;
+  }
+ }
+
+ int ans = 0;
+ for (int i = 29; i >= 0; i--)
+  if (a1[i] == -1)
+  a1[i] = 0;
+ for (int i = 29; i >= 0; i--)
+  ans = ans * 2 + a1[i];
+
+ cout << ans << " ";
+ for (int i = 2; i <= n; i++)
+ {
+  cout << (ans ^ w[i]) << " \n"[i == n];
+ }
+ return;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t;
+ cin >> t;
+ while (t--)
+ {
+ solve();
+ }
+
+ return 0;
+}
+```

docs/ACM/Knowledge/Math/数论/扩展欧几里得.md

@@ -15,4 +15,229 @@ LL exgcd(LL a, LL b, LL &x, LL &y)
 
  return d;
 }
+```
+
+## 牛客第一场 M Water
+
+[题目链接](https://ac.nowcoder.com/acm/contest/57355/M)
+
+[证明链接](https://zhuanlan.zhihu.com/p/644323896)
+
+
+四种操作 两个水杯 容量A、B 凑喝水的量x
+
+
+exgcd uA + vB = x
+
+if(u >= 0 && v >= 0) 直接两个杯子倒喝倒喝就行 ans = 2(u+v)
+
+if(u < 0 || v < 0) 可以构造 ans = 2abs(u)+2abs(v)-1
+
+if(u < 0 && v < 0) 不存在哦
+
+> 严格证明最少这么多操作比较复杂
+
+
+
+```cpp
+#include <bits/stdc++.h>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+LL exgcd(LL a, LL b, LL &x, LL &y)
+{
+ if (b == 0)
+ {
+ x = 1, y = 0;
+ return a;
+ }
+ LL g = exgcd(b, a % b, y, x);
+ y -= a / b * x;
+ return g;
+}
+
+LL get_ans(LL a, LL b)
+{
+ if (a >= 0 && b >= 0)
+ {
+ return 2 * (a + b);
+ }
+
+ return 2 * abs(a) + 2 * abs(b) - 1;
+}
+
+void solve()
+{
+ LL A, B, x;
+ cin >> A >> B >> x;
+
+ if (A > B)
+ swap(A, B);
+
+ LL u, v;
+ LL g = exgcd(A, B, u, v); // uA + vB = g
+
+ A /= g, B /= g;
+ if (x % g != 0)
+ {
+ cout << -1 << endl;
+ return;
+ }
+ x /= g;
+
+ // uA + vB = 1 ==> uA + vB = x
+ u = (u * x % B + B) % B; // 获得大于0的最小的u
+ v = (x - u * A) / B;
+
+ LL ans = 1E18;
+ ans = min(ans, get_ans(u, v));
+ ans = min(ans, get_ans(u - B, v + A));
+
+ v = (v % A + A) % A;
+ u = (x - v * B) / A;
+ ans = min(ans, get_ans(u, v));
+ ans = min(ans, get_ans(u + B, v - A));
+
+ cout << ans << endl;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t;
+ cin >> t;
+ while (t--)
+ {
+ solve();
+ }
+
+ return 0;
+}
+```
+## 杭电第八场 题3 Congruences
+
+[题目链接](https://vjudge.net/problem/HDU-7363)
+
+```cpp
+#include <iostream>
+#include <algorithm>
+#include <cstring>
+#include <vector>
+#include <array>
+#include <numeric>
+
+#define LL long long
+#define ULL unsigned long long
+#define x first
+#define y second
+#define endl "\n"
+
+using namespace std;
+
+typedef pair<int, int> PII;
+typedef pair<LL, LL> PLL;
+
+const int INF = 0x3f3f3f3f;
+
+LL qmi(LL a, LL k, LL p)
+{
+ if (a == 0)
+ return 0;
+ LL res = 1;
+ while (k)
+ {
+ if (k & 1)
+ res = (__int128)res * a % p;
+ a = (__int128)a * a % p;
+ k >>= 1;
+ }
+ return res;
+}
+
+LL exgcd(LL a, LL b, LL &x, LL &y)
+{
+ if (b == 0)
+ {
+ x = 1, y = 0;
+ return a;
+ }
+
+ LL g = exgcd(b, a % b, y, x);
+ y -= (__int128)a / b * x;
+
+ return g;
+}
+
+void solve()
+{
+ LL n, m;
+ cin >> m;
+
+ vector<LL> p(m + 1), q(m + 1);
+ LL M = 1;
+ for (int i = 0; i < m; i++)
+ {
+  cin >> p[i] >> q[i];
+  M *= p[i];
+ }
+
+ LL ans = 0;
+ for (int i = 0; i < m; i++)
+ {
+  LL Mi = M / p[i];
+  LL ti, x;
+  exgcd(Mi, p[i], ti, x);
+  ans = ((__int128)ans + (__int128)q[i] * Mi * ti % M) % M;
+ }
+
+ ans = (ans % M + M) % M;
+
+ bool ok = true;
+ for (int i = 0; i < m; i++)
+ {
+  LL lhs = qmi(ans, p[i], M);
+  if (lhs != q[i])
+  {
+  ok = false;
+  break;
+  }
+ }
+
+ if (ans == 0)
+  ans = M;
+
+ if (!ok)
+  ans = -1;
+ cout << ans << endl;
+
+ return;
+}
+
+int main()
+{
+ ios::sync_with_stdio(false);
+ cin.tie(0), cout.tie(0);
+
+ int t;
+ cin >> t;
+ while (t--)
+ {
+  solve();
+ }
+
+ return 0;
+}
 ```