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");
        }
    }

}