题目描述
代码展示
#include <iostream>
using namespace std;
const int maxn = 100005;
int n, q, x, a[maxn];
int main() {
scanf("%d%d", &n, &q);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
while (q--) {
scanf("%d", &x);
int l = 0, r = n - 1;
while (l < r) {
int mid = l + r >> 1;
if (a[mid] < x) l = mid + 1;
else r = mid;
}
if (a[l] != x) {
printf("-1 -1\n");
continue;
}
int l1 = l, r1 = n;
while (l1 + 1 < r1) {
int mid = l1 + r1 >> 1;
if (a[mid] <= x) l1 = mid;
else r1 = mid;
}
printf("%d %d\n", l, l1);
}
return 0;
}
AcWing 789. 数的范围(详细分析二分过程)