Problem: 蓝桥杯 递增三元组

思路

这是一个关于数组的问题，我们需要找到一个递增的三元组。这个三元组由三个数组中的元素组成，每个数组提供一个元素，并且这三个元素满足递增的关系。

解题方法

我们可以使用三种方法来解决这个问题：前缀和，二分查找和双指针。
1、前缀和：我们首先对数组a和c进行计数，然后计算出前缀和。然后，我们遍历数组b，对于每一个元素bi，我们找到数组a中小于bi的元素数量和数组c中大于bi的元素数量，然后将这两个数量相乘，累加到结果中。
2、二分查找：我们首先对数组a和c进行排序。然后，我们遍历数组b，对于每一个元素bi，我们使用二分查找在数组a中找到小于bi的最大元素的位置，和在数组c中找到大于bi的最小元素的位置，然后将这两个位置相乘，累加到结果中。
3、双指针：我们首先对数组a，b和c进行排序。然后，我们使用两个指针，一个指向数组a的开始，一个指向数组c的开始。我们遍历数组b，对于每一个元素bi，我们移动数组a的指针，直到找到一个大于或等于bi的元素，然后移动数组c的指针，直到找到一个大于bi的元素，然后将这两个位置相乘，累加到结果中。

复杂度

时间复杂度:

对于前缀和和双指针方法，时间复杂度为$O(n)$。对于二分查找方法，时间复杂度为$O(nlogn)$。

空间复杂度:

对于所有的方法，空间复杂度都为$O(n)$。

前缀和Code

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;public class Main {static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));static StreamTokenizer sr = new StreamTokenizer(in);static int MAXN = (int) (1e5 + 10);static int[] a = new int[MAXN];static int[] b = new int[MAXN];static int[] c = new int[MAXN];static int[] as = new int[MAXN];static int[] cs = new int[MAXN];static int n;public static void main(String[] args) throws IOException {n = nextInt();for (int i = 0; i < n; i++) {a[i] = nextInt();}for (int i = 0; i < n; i++) {b[i] = nextInt();}for (int i = 0; i < n; i++) {c[i] = nextInt();}for (int i = 0; i < n; i++) {as[a[i]]++;cs[c[i]]++;}// 计算出前缀和for (int i = 1; i < MAXN; i++) {as[i] += as[i - 1];cs[i] += cs[i - 1];}long res = 0;// 枚举枚举每一个bifor (int i = 0; i < n; i++) {if (b[i] == 0) {continue;}res += (long) as[b[i] - 1] * (long) (cs[MAXN - 1] - cs[b[i]]);}out.println(res);out.flush();}static int nextInt() throws IOException {sr.nextToken();return (int) sr.nval;}}

二分Code

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;public class Main {static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));static StreamTokenizer sr = new StreamTokenizer(in);static int MAXN = (int) (1e5 + 10);static int[] a = new int[MAXN];static int[] b = new int[MAXN];static int[] c = new int[MAXN];static int n;public static void main(String[] args) throws IOException {n = nextInt();for (int i = 0; i < n; i++) {a[i] = nextInt();}for (int i = 0; i < n; i++) {b[i] = nextInt();}for (int i = 0; i < n; i++) {c[i] = nextInt();}Arrays.sort(a, 0, n);Arrays.sort(c, 0, n);// 二分查找 小于bi最右边的数字 目的尽可能多// 大于bi最左边的数字 目的是让数字尽可能多long res = 0;for (int i = 0; i < n; i++) {int left = getA(0, n - 1, b[i]);int right = getB(0, n - 1, b[i]);res += (long) (left + 1) * (n - right);}out.println(res);out.flush();}private static int getA(int l, int r, int x) {// TODO Auto-generated method stubwhile (l < r) {int mid = (l + r + 1) / 2;if (a[mid] < x) {l = mid;} else {r = mid - 1;}}return a[l] < x ? l : -1;}private static int getB(int l, int r, int x) {// TODO Auto-generated method stubwhile (l < r) {int mid = (l + r) / 2;if (c[mid] > x) {r = mid;} else {l = mid + 1;}}return c[r] > x ? r : n;}static int nextInt() throws IOException {sr.nextToken();return (int) sr.nval;}}

双指针Code

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;public class Main {static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));static StreamTokenizer sr = new StreamTokenizer(in);static int MAXN = (int) (1e5 + 10);static int[] a = new int[MAXN];static int[] b = new int[MAXN];static int[] c = new int[MAXN];static int n;public static void main(String[] args) throws IOException {n = nextInt();for (int i = 0; i < n; i++) {a[i] = nextInt();}for (int i = 0; i < n; i++) {b[i] = nextInt();}for (int i = 0; i < n; i++) {c[i] = nextInt();}Arrays.sort(a, 0, n);Arrays.sort(b, 0, n);Arrays.sort(c, 0, n);// 使用双指针算法long res = 0;for (int i = 0, ai = 0, ci = 0; i < n; i++) {while (ai < n && a[ai] < b[i]) {ai++;}while (ci < n && c[ci] <= b[i]) {ci++;}res += (long) ai * (n - ci);}out.println(res);out.flush();}static int nextInt() throws IOException {sr.nextToken();return (int) sr.nval;}}