如题，参考一个C++可以过的全部TLE

这个是80分的```c #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;

}

这个是C++可以过的
```cpp
#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;
}

参考上面用c的全部tle

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

int main(void) {
	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]);
	}
	while (m--) {
		int num;
		scanf("%d", &num);
		int right = n;
		int lift = 1;
		while (lift < right) {
			int mid = lift + (1 + right ) / 2;
			if (M[mid] >= num) {
				right = mid;
			} else {
				lift = mid + 1;
			}
		}
		if (M[lift] == num) {
			printf("%d", num);
		} else {
			printf("-1");
		}
	}


}