分治方法自动查找！

#include <iostream>
#include <cstdio>
using namespace std;
int a[1000005];
int x;
int check1(int left, int right){//节省了很多代码
	if(a[left] == x) return left;
	if(a[left] > x || a[right] < x) return 0x3f3f3f3f;
	return 0;
}int check2(int left, int right){
	if(a[right] == x) return right;
	if(a[left] > x || a[right] < x) return -1;
	return 0;
}int check3(int left, int right){
	if(a[left] > x) return left;
	if(a[right] <= x) return 0x3f3f3f3f;
	return 0;
}
int find(int left, int right, int check(int, int), bool is_up = 0){//这里只需要填check函数，是否要最大值就行了
	int cxk = check(left, right);
	if(cxk) return cxk;
	if(left == right) return left;
	int mid = (left + right) / 2;
	if(is_up) return max(find(left, mid, check, 1), find(mid + 1, right, check, 1));
	return min(find(left, mid, check), find(mid + 1, right, check));
}
int main(){
	int n, m;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%d", a + i);
	}scanf("%d", &m);
	while(m--){
		scanf("%d", &x);
		int f1 = find(1, n, check1), f2 = find(1, n, check2, 1), f3 = find(1, n, check3);
		printf("%d %d %d\n", (f1 == 0x3f3f3f3f ? -1 : f1), (f2 == 0x3f3f3f3f ? -1 : f2), (f3 == 0x3f3f3f3f ? -1 : f3));
	}
	

	return 0;
}

单次查找时间复杂度：$只要check函数严谨就是O(log_2n)$