upper_bound(a,a+n,k)函数求出的值就表示的是指向满足ai>k的ai的最小指针

Question

You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).

Output

Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.

Example

Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4

---

Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5

Code

#include 
#include 
using namespace std;

int a[200005],b[200005];
int bin_search(int a[],int r,int l,int v);
int main()
{
    int n,m;int i;
    int l,r,mid=0;
    cin>>n>>m;
    for(i=0;icin>>a[i];
    sort(a,a+n);
    for(i=0;icin>>b[i];
    cout<0,n,b[0]);
    for(i=1;icout<<' '<0,n,b[i]);
    }
    cout<return 0;
}
int bin_search(int a[],int l,int r,int v)
{
    int mid;
    while(l2;
        if(a[mid]<=v)
            l=mid+1;
        else if(a[mid]>v)
            r=mid;
    }
    return l;
}

事实上不必这么麻烦。C++STL里面的upper_bound(a,a+n,k)函数求出的值就表示的是指向满足ai>k的ai的最小指针。
所以用STL求解很简单。

#include 
#include 
using namespace std;
int a[200005],b[200005];

int main()
{
    int n,m;int i;
    int l,r,mid=0;
    cin>>n>>m;
    for(i=0;icin>>a[i];
    sort(a,a+n);
    for(i=0;icin>>b[i];
    cout<0])-a;//与首地址做差
    for(i=1;icout<<' '<cout<return 0;
}