前言

整体评价

这场牛客周赛很特别，没有一题是水题（签到），也没真正意义上的压轴题。

牛客周赛还是注重数学，B，D是数学题，A是模拟题，C的解法偏多，滑窗/26组前缀和&二分皆可。

A. 小红的环形字符串

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();char[] str = sc.next().toCharArray();// 等价原理// 1. 先当stack使用int ans = 0;Deque<Character> deq = new ArrayDeque<>();for (int i = 0; i < n; i++) {char c = str[i];if (!deq.isEmpty() && deq.peekLast() == c) {ans += 2;deq.pollLast();} else {deq.addLast(c);}}// 2. 双端队列，合并头和尾while (deq.size() >= 2 && deq.peekFirst() == deq.peekLast().charValue()) {ans += 2;deq.pollFirst();deq.pollLast();}System.out.println(ans);}}

B. 小红的乘除变换

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int t = sc.nextInt();while (t-- > 0) {long x = sc.nextLong();long y = sc.nextLong();int ny5 = 0, ny6 = 0;while (y > 0 && y % 5 == 0) { y /= 5; ny5++; }while (y > 0 && y % 6 == 0) { y /= 6; ny6++; }int nx5 = 0, nx6 = 0;while (x > 0 && x % 5 == 0) { x /= 5; nx5++; }while (x > 0 && x % 6 == 0) { x /= 6; nx6++; }if (x != y || nx5 > ny5 || nx6 < ny6) {System.out.println(-1);} else {System.out.println(ny5 - nx5 + nx6 - ny6);}}}}

C. 小红的子串

滑窗

滑窗可以用三指针，也可以用双指针

三指针，枚举右(左)端点, 然后滑窗维护[l, r]不同字符种类数的边界p1, p2, 结果为

ans += abs(p2 - p1)

双指针，定义f(x)为种类数 >= x的方案总数

res = f(l) - f(r + 1)

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int l = sc.nextInt(), r = sc.nextInt();char[] str = sc.next().toCharArray();Map<Integer, Integer> hash1 = new HashMap<>();Map<Integer, Integer> hash2 = new HashMap<>();long ans = 0;int p1 = 0, p2 = 0;for (int i = 0; i < n; i++) {int v = str[i] - 'a';hash1.merge(v, 1, Integer::sum);hash2.merge(v, 1, Integer::sum);// 维护有r种不同字符的边界, > rwhile (p1 <= i && hash1.size() > r) {hash1.computeIfPresent(str[p1] - 'a', (a, b) -> b > 1 ? b - 1 : null);p1++;}// 维护有l种不同字符的边界, >=lwhile (p2 <= i && hash2.size() >= l) {hash2.computeIfPresent(str[p2] - 'a', (a, b) -> b > 1 ? b - 1 : null);p2++;}ans += (p2 - p1);}System.out.println(ans);}}

import java.io.*;
import java.util.*;public class Main {/*双指针，定义f(x)为种类数 >= x的方案总数res = f(l) - f(r + 1) */static long solve(char[] str, int x) {Map<Integer, Integer> hash = new HashMap<>();long ans = 0;int j = 0;for (int i = 0; i < str.length; i++) {int v = str[i] - 'a';hash.merge(v, 1, Integer::sum);while (j <= i && hash.size() >= x) {hash.computeIfPresent(str[j] - 'a', (a, b) -> b > 1 ? b - 1 : null);j++;}ans += j;}return ans;}public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int l = sc.nextInt(), r = sc.nextInt();char[] str = sc.next().toCharArray();System.out.println(solve(str, l) - solve(str, r + 1));}}

前缀和&二分

字符种类最多是26，是有限可枚举

因此这也是一个切入点

构建26 slots的前缀和

然后求二分求 [l, r]的两个边界

import java.io.*;
import java.util.*;public class Main {static int binarySearch(int[][] pre, int e, int x) {int l = 0, r = e;while (l <= r) {int m = l + (r - l) / 2;// 统计不同字符种类个数int cnt = 0;for (int i = 0; i < 26; i++) {if (pre[i][e + 1] - pre[i][m] > 0) cnt++;}// cnt >= x 是 check核心逻辑if (cnt >= x) {l = m + 1;} else {r = m - 1;}}return r;}public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int n = sc.nextInt();int l = sc.nextInt(), r = sc.nextInt();char[] str = sc.next().toCharArray();int[][] pre = new int[26][n + 1];// 构建26个前缀和for (int i = 0; i < n; i++) {for (int j = 0; j < 26; j++) {pre[j][i + 1] = pre[j][i] + (str[i] - 'a' == j ? 1 : 0);}}long ans = 0;for (int i = 0; i < n; i++) {ans += binarySearch(pre, i, l) - binarySearch(pre, i, r + 1);}System.out.println(ans);}}

D. 回文权值和

import java.io.*;
import java.util.*;public class Main {public static long ksm(long b, long v, long mod) {long r = 1l;b = b % mod;while (v > 0) {if (v % 2 == 1) r = r * b % mod;b = b * b % mod;v /= 2;}return r;}public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));long n = sc.nextLong();// n < 3没有解if (n < 3) {System.out.println(0);return;}long mod = (long)1e9 + 7;long res = 676l * ksm(26, n - 3, mod) % mod * ((n - 2) % mod) % mod;System.out.println(res);}}

写在最后