[c language] limit the number of searches and output the maximum value found in the number of internal searches

1、 subject

Give you a length of n Array of A. Ask you m Time ： In the array A in , Less than or equal to a given number xi What is the maximum value of ？

Input format

• first line ： Two integers n and m, Represents the length of the array and the number of lookups .
• The second line ：n It's an integer ai, Represents each number in the array .
• Next m That's ok , Each row has 1 It's an integer xi , Integer representing lookup .

Output format

• For each query , If you can find , The output is less than or equal to xi The maximum of .
• Otherwise output -1.

Examples ：

Input ：
10 4
1 2 3 3 3 5 5 7 7 9
0
4
9
10
Output ：
-1
3
9
9

2、 Complete code

#include <iostream>
#include <algorithm>
using namespace std;

int n, m;
int a[100001];

int judge(int input,int a[]) {
    
	int res;
	//  Query cannot find direct output -1
	if (input < a[0]) {
    
		res = -1;
	}
	//  If the number of lookups is longer than the array , Output the last number of sorted array directly 
	else if (input >= a[n - 1]) {
    
		res = a[n - 1];
	}
	else if (input == a[0]) {
    
		res = a[0];
	}
	//  Two points search 
	int ans = -1;
	int left = 0, right = n - 1;
	while (left <= right) {
    
		//  Moves to the right one 
		int mid = (left + right) >> 1;
		if (a[mid] <= input) {
    
			ans = a[mid];
			left = mid + 1;
		}
		else {
    
			right = mid - 1;
		}
	}
	res = ans;
	return res;
}

int main() {
    

	//  Array length and number of lookups 
	cin >> n >> m;
	//  Enter full array 
	for (int i = 0; i < n; i++) {
    
		cin >> a[i];
	}
	//  Array sorting , Default ascending order 
	sort(a, a + n);
	for (int i = 0; i < m; i++) {
    
		int x;
		//  Enter the number you want to find 
		cin >> x;
		cout << judge(x, a) << endl;
	}
	return 0;
}

3、 Screenshot