O(2e5 + n + m)，不愧是我（

#include <iostream>
#include <cstdio>
using namespace std;
int a[100005], xiabiao[200005], ans[100005];
int n, m, x, ct;
void _init(){
	for(int i = 1; i <= 200005; i++){
		xiabiao[i] = 0x3f3f3f3f;
	}
	for(int i = 1; i <= n; i++){
		xiabiao[a[i] + 100000] = i;
	}
}
int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++){
		scanf("%d", &a[i]);
	}_init();
	while(m--){
		scanf("%d", &x);
		if(xiabiao[x + 100000] == 0x3f3f3f3f) ans[++ct] = -1;
		else ans[++ct] = xiabiao[x + 100000];
	}for(int i = 1; i <= ct; i++){
		printf("%d\n", ans[i]);
	}

	return 0;
}