【整数二分】Acwing789.数的范围

Acwing789.数的范围

题解

最左边界,每次 mid >= x 则 r = mid, 再找左边是否有符合要求的边界。
最右边界,每次 mid <= x 则 l = mid, 再找右边是否有符合要求的边界。（注意当l = r-1时 ，（l+r）>> 1 == l 故我们需要 mid = （l + r）>> 1 + 1 ）

写法一：多加一个else(易懂)

#include <iostream>
#include <cstdio>

using namespace std;

int arr[100002];
int n,q;
int k;

int find_left()
{
    int mid, bg = 0, ed = n-1;
    int res = -1;
    while(bg <= ed){
        mid = (bg+ed)/2;
        if(arr[mid] < k)
                bg = mid+1;
        else if(arr[mid] > k)
            ed = mid-1;
        else
            res = mid,ed = mid-1;
    }
    return res;
}

int find_right()
{
    int mid,bg = 0, ed = n-1;
    int res = -1;
    
    while(bg <= ed)
    {
        mid = (bg+ed)/2;
        if(arr[mid] < k)
                bg = mid+1;
        else if(arr[mid] > k)
            ed = mid-1;
        else
            res = mid,bg = mid+1;
    }
    return res;
}

int main()
{
    scanf("%d%d",&n,&q);
    for(int i = 0; i < n; ++i)
        scanf("%d",&arr[i]);
    while(q--){
      scanf("%d",&k);
      printf("%d %d\n",find_left(),find_right());
    }
    return 0;
}

写法二：简略

#include <iostream>
#include <cstdio>

using namespace std;

const int N = 100010;

int a[N];

int binary_left(int l, int r, int k)
{
    int mid;
    while( l < r)
    {
        mid = l + r >> 1;
        if(a[mid] >= k) r = mid;
        else l = mid + 1;
    }
    return l;
}

int binary_right(int l, int r, int k)
{
    int mid;
    while( l < r )
    {
        mid = (l + r >> 1) + 1;
        if(a[mid] <= k) l = mid;
        else r = mid - 1;
    }
    return l;
}

int main()
{
    int n, q;
    scanf("%d%d",&n, &q);
    for(int i = 0; i < n; ++i)
        scanf("%d",&a[i]);
    int k, l, r;
    while(q--)
    {
        scanf("%d",&k);
        l = binary_left(0, n-1, k);
        r = binary_right(0, n-1, k);
        if(a[l] != k)
            printf("-1 -1\n");
        else
            printf("%d %d\n",l,r);
    }
    return 0;
}