vjudge, ADS, KBTU, Rabin-Karp and Knuth-Morris-Pratt Algorithms

Author: zotvent

vjudge/ADS, KBTU, Rabin-Karp and Knuth-Morris-Pratt Algorithms/A - Ada and Spring Cleaning/A - Ada and Spring Cleaning.cpp

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+#include <stdio.h>
+#include <iostream>
+#include <fstream>
+#include <vector>
+#include <algorithm>
+#include <queue>
+#include <deque>
+#include <stack>
+#include <set>
+#include <unordered_set>
+#include <map>
+#include <unordered_map>
+#include <cmath>
+#include <cstring>
+#include <bits/stdc++.h>
+
+using namespace std;
+
+#define F first
+#define S second
+#define pb push_back
+#define pf push_front
+#define ppb pop_back
+#define ppf pop_front
+#define mp make_pair
+#define speed ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
+
+typedef long long ll;
+typedef unsigned long long ull;
+typedef pair<int, int> pii;
+
+const int inf = 1e9;
+const int N = 1e5+5;
+const int mod = 1e9 + 7;
+const int MAX = 1e6;
+const int shift = 4;
+const int p[] = {31, 53};
+
+class Solution {
+    int t, n, k;
+    string text;
+    set<pair<ull, ull>> unique;
+    vector<ull> p_pow[2], hash[2];
+    
+    void print() {
+        cout << unique.size() << endl;
+    }
+    
+    void input() {
+        speed
+        // freopen("input.txt", "r", stdin);
+        // freopen("output.txt", "w", stdout);
+        cin >> t;
+    }
+    
+    void output() {
+    }
+    
+    void solution() {
+        calc_pows(N);
+        for (int j = 0; j < t; j++) {
+            cin >> n >> k >> text;
+            prepare();
+
+            for (int i = 0; i < n - k + 1; i++) {
+                ull cur[] = {hash[0][i + k - 1], hash[1][i + k - 1]};
+                if (i) {
+                    cur[0] -= hash[0][i - 1];
+                    cur[1] -= hash[1][i - 1];                    
+                }
+                cur[0] *= p_pow[0][n - i - 1];
+                cur[1] *= p_pow[1][n - i - 1];
+                unique.insert(mp(cur[0], cur[1]));
+            }
+
+            print();
+            unique.clear();
+        }        
+    }
+
+    void calc_pows(int k) {
+        // all pows of p
+        p_pow[0].resize(k);
+        p_pow[1].resize(k);
+        
+        p_pow[0][0] = 1;
+        p_pow[1][0] = 1;
+
+        for (int i = 1; i < k; i++) {
+            p_pow[0][i] = p_pow[0][i - 1] * p[0];
+            p_pow[1][i] = p_pow[1][i - 1] * p[1];
+        }
+    }
+
+    void prepare() {
+        // all hashes of text
+        hash[0].resize(n);
+        hash[1].resize(n);
+        for (int i = 0; i < n; i++) {
+            hash[0][i] = (text[i] - 'a' + 1) * p_pow[0][i];
+            hash[1][i] = (text[i] - 'a' + 1) * p_pow[1][i];
+            if (i) {
+                hash[0][i] += hash[0][i - 1];
+                hash[1][i] += hash[1][i - 1];
+            }
+        }
+    }
+    
+public:
+    
+    Solution() {
+    }
+    
+    void solve() {
+        input();
+        solution();
+        output();
+    }
+};
+
+int main() {
+    Solution s = Solution();
+    s.solve();
+    return 0;
+}

vjudge/ADS, KBTU, Rabin-Karp and Knuth-Morris-Pratt Algorithms/A - Ada and Spring Cleaning/task.md

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+# Task
+
+Ada the Ladybug has decided to do some "Spring Cleaning". As you might know, she keeps a TODO list. She is very sparing so she keeps all her activities as one string. You might get very confused while reading the string but she has a system - every activity has length exactly K characters. Sadly, as new activities were added to the list many duplicities appeared. Now it is time to find out how many unique activities are in her TODO list.
+
+# Input
+
+First line contains T, number of test-cases.
+
+Each test-case begins with N, K, 1 ≤ K ≤ N ≤ 105, length of string and length of activites respectively.
+
+Next line consists of string of length N, consisting of lowercase letters.
+
+The sum of lengths of strings among all test-cases won't exceed 3 * 1e5
+
+# Output
+
+For each test-case, print the number of unique substrings of length K
+
+# Examples
+
+## Input
+
+5
+3 2
+aaa
+5 1
+abcba
+4 2
+abac
+10 2
+abbaaaabba
+7 3
+dogodog
+
+## Output
+
+1
+3
+3
+4
+4