题目描述
找x,
如果存在,输出所求元素的起始位置和终止位置
如果不存在,输出-1 -1
算法1
(y总强大的模板) $O(logN)$
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int q[N];
int main()
{
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) scanf("%d", &q[i]);
while (m--)
{
int x;
cin >> x;
int l = 0, r = n - 1;
while (l < r)
{
int mid = l + r >> 1;
if (q[mid] >= x) r = mid; //对于3...3 要找到的mid都要大于等于3,然后往左找左边界
else l = mid +1;
}
if (q[l] != x)
{
cout << "-1 -1" << endl;
continue;
}
cout << l << ' ';
l = 0, r = n - 1;
while (l < r)
{
int mid = l + r + 1 >> 1;
if (q[mid] <= x) l = mid; ////对于3...3 找到的mid都要小于等于3, 往右找右边界
else r = mid - 1;
}
cout << l << endl;
}
return 0;
}
算法2
(STL函数) $O(logN)$ ??
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
int q[N];
int main()
{
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) scanf("%d", &q[i]);
while (m--)
{
int x;
cin >> x;
if (!binary_search(q, q + n, x))
cout << "-1 -1" << endl;
else
cout << lower_bound(q, q + n, x) - q<< ' ' << upper_bound(q, q + n, x) - q - 1 << endl;
}
return 0;
}