一、視頻講解

[藍橋杯]真題講解：抓娃娃（思維+二分））

二、正解代碼

1、C++

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef pair<int,int> pii;
const int N = 1e5 + 10;
int n, m;int main()
{cin >> n >> m;vector<pii>a(n);vector<double>b(m);for(int i = 0; i < n; i ++) {int L, R; cin >> L >> R;double mid = (L + R) / 2.0;b[i] = mid;}for(int i = 0; i < m; i ++) {cin >> a[i].x >> a[i].y;}sort(b.begin(), b.end());auto check1 = [&](int mid, int i) {return b[mid] >= a[i].x;};auto check2 = [&](int mid, int i) {return b[mid] <= a[i].y;};for(int i = 0; i < m; i ++) {int L = a[i].x, R = a[i].y;int l1 = 0, r1 = n - 1;while(l1 < r1) {int mid = l1 + r1 >> 1;if(check1(mid, i))r1 = mid;elsel1 = mid + 1;}int l2 = 0, r2 = n - 1;while(l2 < r2) {int mid = l2 + r2 + 1 >> 1;if(check2(mid, i))l2 = mid;elser2 = mid - 1;}if(b[l1] >= L && b[l2] <= R) {cout << l2 - l1 + 1 << endl;}else{cout << 0 << endl;}}return 0;
}

2、python3

n, m = map(int, input().split())
tmp = [list(map(int, input().split())) for _ in range(n)]
b = []
for x, y in tmp:b.append ((x + y) / 2.0)a = [list(map(int, input().split())) for _ in range(m)]b.sort()def check1(mid: int, i: int)->bool:return b[mid] >= a[i][0]def check2(mid: int, i: int)->bool:return b[mid] <= a[i][1]for i in range(m):L, R = a[i]l1, r1 = 0, n - 1while l1 < r1:mid = (l1 + r1) // 2if check1(mid, i):r1 = midelse:l1 = mid + 1l2, r2 = 0, n - 1while l2 < r2:mid = (l2 + r2 + 1) // 2if check2(mid, i):l2 = midelse:r2 = mid - 1if b[l1] >= L and b[l2] <= R:print(l2 - l1 + 1)else:print(0)

3、Java

import java.util.Arrays;
import java.util.Scanner;public class Main {public static int n, m;public static double[] b;public static int[][] a;public static void main(String[] args) {Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();b = new double[n];a = new int[n][2];for(int i = 0; i < n; i ++) {int L = sc.nextInt();int R = sc.nextInt();double mid = (L + R) / 2.0;b[i] = mid;}for(int i = 0; i < m; i ++) {a[i][0] = sc.nextInt();a[i][1] = sc.nextInt();}Arrays.sort(b);for(int i = 0; i < m; i ++) {int L = a[i][0], R = a[i][1];int l1 = 0, r1 = n - 1;while(l1 < r1){int mid = l1 + r1 >> 1;if(check1(mid, i)) {r1 = mid;}else{l1 = mid + 1;}}int l2 = 0, r2 = n - 1;while(l2 < r2) {int mid = l2 + r2 + 1 >> 1;if(check2(mid, i)){l2 = mid;}else{r2 = mid - 1;}}if(b[l1] >= L && b[l2] <= R) {System.out.println(l2 - l1 + 1);}else{System.out.println(0);}}}private static boolean check1(int mid, int i) {return b[mid] >= a[i][0];}private static boolean check2(int mid, int i){return b[mid] <= a[i][1];}
}