Given an integer array of length n arranged in ascending order, and q queries.
For each query, return the starting position and ending position of an element k (the position starts counting from 0).
If the element does not exist in the array, "-1 -1" is returned.
Input format The
first line contains integers n and q, indicating the length of the array and the number of queries.
The second line contains n integers (all in the range of 1 to 10000), representing a complete array.
Next q lines, each line contains an integer k, which represents a query element.
Output format
A total of q lines, each line contains two integers, indicating the starting position and ending position of the element to be sought.
If the element does not exist in the array, "-1 -1" is returned.
Data range
1≤n≤100000
1≤q≤10000
1≤k≤10000
Input example:
6 3
1 2 2 3 3 4
3
4
5
Output example:
3 4
5 5
-1 -1
AC code:
#include<stdio.h>
#include<algorithm>
int a[100010];
int n,q;
int head,tail;
void binary(int x)
{
int l=-1,r=n;
int middle;
head=n-1;
tail=0;
while(l+1!=r)//找第一个大于等于x的位置
{
//求最小，向下取整
middle=l+((r-l)>>1);
if(a[middle]<x){
l=middle;
}
else if(a[middle]>=x){
r=middle;
if(a[middle]==x) head=std::min(head,middle);
}
}
if(a[head]==x){
//x存在则继续找第一个小于等于x的位置
l=middle-1;
r=n;
while(l+1!=r)
{
//求最大，向上取整
middle=l+((r-l+1)>>1);
if(a[middle]<=x){
l=middle;
if(a[middle]==x) tail=std::max(tail,middle);
}
else if(a[middle]>x){
r=middle;
}
}
}
else {
//x不存在
head=-1;
tail=-1;
}
}
int main()
{
scanf("%d%d",&n,&q);
for(int i=0;i<n;++i)
{
scanf("%d",&a[i]);
}
for(int i=0;i<q;++i)
{
int x;
scanf("%d",&x);
binary(x);
printf("%d %d\n",head,tail);
}
return 0;
}