64分，样例过，求解！

Bramble_Marshall @ 2024-07-16 08:01:13

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;

int a[N];

int n, m;

int low_find(int num, int l, int r) {
    int mid = l + (r - l) / 2;
    if (l > r) return -1;
    else if (a[mid] > num) return low_find(num, l, mid - 1);
    else if (a[mid] < num) return low_find(num, mid + 1, r);
    else while (a[mid] == num) mid--;
    return ++mid;
}

// 13489 : 6

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1, tmp; i <= m; i++) {
        cin >> tmp;
        cout << low_find(tmp, 1, n) << ' ';
    }
    return 0;
}

by Youth_Glory @ 2024-07-16 08:19:00

可能是二分判断有点问题吧，你可以试着稍微改改

by eleke @ 2024-07-16 09:05:00

直接用二分函数

#include <iostream>
#include <algorithm>
using namespace std;
int a[1000010];

int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= m; i++)
    {
        int t;
        cin >> t;
        int pos = lower_bound(a + 1, a + n + 1, t) - a;
            //lower_bound(a + 1, a + n + 1, t);
            //查找a数组中第一个大于等于t的数的位置
        if (t == a[pos]) cout << pos << " ";
        else cout << -1 << " ";
    }
    return 0;
}