这个题可以用C吗，我自己写的二分最后一个超时

#include <stdio.h>
#include <string.h>
int M[1000050];

int main() {
	int n, m; //m是查询次数
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &M[i]);
	}
	for (int i = 0; i < m; i++) {
		int num;
		scanf("%d", &num);
		int relust = -1;
		int right = n;
		int lift = 1;
		while (lift <= right) {
			int mid = (lift + right ) / 2;
			if (num == M[mid]) {
				relust = mid;
				for (int i = mid - 1; i >= lift; i--) {
					if (num == M[i]) {
						relust = i;
					}
				}
				break;
			} else if (num < M[mid]) {
				right = mid - 1;
			} else if (num > M[mid]) {
				lift = mid + 1;
			}
		}

		printf("%d ", relust);
	}
	return 0;

}

参考一个C++可以过的全部TLE

#include<bits/stdc++.h>

using namespace std;
const int N = 1000010;
int n, m;
int q[N];
int main(){
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++){
        scanf ("%d", &q[i]); 
    }
    while(m--) {
        int x;
        scanf("%d", &x);
        
        int l = 1, r = n;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (q[mid] >= x) r = mid;
            else l = mid + 1;
        }
        
        if (q[l] == x) printf("%d ", l);
        else printf("-1 ");
    }
     
    return 0;
}