From 34a0fcd040a8c8109429828f49805b5eb3cd554e Mon Sep 17 00:00:00 2001
From: Ariel <ariel@arielherself.xyz>
Date: Sun, 11 Feb 2024 12:36:03 +0800
Subject: [PATCH] Create prefix-hash.cc

---
 string/prefix-hash.cc | 43 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 string/prefix-hash.cc

diff --git a/string/prefix-hash.cc b/string/prefix-hash.cc
new file mode 100644
index 0000000..8d59c3e
--- /dev/null
+++ b/string/prefix-hash.cc
@@ -0,0 +1,43 @@
+class Solution {
+public:
+    int countMatchingSubarrays(vector<int>& nums, vector<int>& pattern) {
+        constexpr int b = 31;
+        int n = nums.size(), m = pattern.size();
+        ll hp1 = 0, hp2 = 0;
+        ll pow1 = 1, pow2 = 1;
+        for (int i = 1; i <= m; ++i) {
+            hp1 = mod(hp1 + mod(pow1 * (pattern[i-1] + 1), MDL1), MDL1);
+            hp2 = mod(hp2 + mod(pow2 * (pattern[i-1] + 1), MDL2), MDL2);
+            pow1 = mod(pow1 * b, MDL1);
+            pow2 = mod(pow2 * b, MDL2);
+        }
+        vector<ll> hn1(n + 1), hn2(n + 1);
+        pow1 = 1, pow2 = 1;
+        for (int i = 2; i <= n; ++i) {
+            int p;
+            if (nums[i-1] > nums[i-2]) p = 1;
+            else if (nums[i-1] == nums[i-2]) p = 0;
+            else p = -1;
+            hn1[i] = mod(hn1[i-1] + mod(pow1 * (p + 1), MDL1), MDL1);
+            hn2[i] = mod(hn2[i-1] + mod(pow2 * (p + 1), MDL2), MDL2);
+            pow1 = mod(pow1 * b, MDL1);
+            pow2 = mod(pow2 * b, MDL2);
+        }
+        int res = 0;
+        pow1 = 1, pow2 = 1;
+        for (int i = 1; i + m <= n; ++i) {
+            if (mod(hp1 * pow1, MDL1) == mod(hn1[i+m] - hn1[i], MDL1) &&
+                mod(hp2 * pow2, MDL2) == mod(hn2[i+m] - hn2[i], MDL2)) {
+                res += 1;
+            }
+            pow1 = mod(pow1 * b, MDL1);
+            pow2 = mod(pow2 * b, MDL2);
+        }
+        return res;
+    }
+};
+
+作者：subcrip
+链接：https://leetcode.cn/problems/number-of-subarrays-that-match-a-pattern-ii/solutions/2637701/zi-fu-chuan-ha-xi-by-subcrip-r3z4/
+来源：力扣（LeetCode）
+著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。