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【acwing】789. 数的范围(二分基础)
2022-07-07 07:46:00 【percation】
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int a[N];
int n,q;
int main(){
scanf("%d%d",&n,&q);
for(int i = 0; i < n; i++){
scanf("%d",&a[i]);
}
while(q--){
int x;
scanf("%d",&x);
int l = 0, r = n - 1;
while(l < r){
int mid = l + r >> 1;
if(a[mid] >= x){
r = mid;
}
else{
l = mid + 1;
}
}
if(a[l] == x) cout << l <<" ";//当前指针所指向的元素是否等于所查询的元素,若等于则说明该元素存在,返回该元素的左端点
else{
cout << "-1 -1" << endl;
continue;
}
l = 0, r = n - 1;//判断所查询元素的右端点的位置
while(l < r){
int mid = (l + r + 1)>> 1;
if(a[mid] <= x){
l = mid;
}
else{
r = mid - 1;
}
}
cout << l << endl;
}
return 0;
}
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