Codeforces 600B Queries about less or equal elements(二分查找)

B. Queries about less or equal elements

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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 ≤ a_i ≤ 10⁹).

The third line contains m integers — the elements of array b ( - 10⁹ ≤ b_j ≤ 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 b_j.

Sample test(s)

input

5 41 3 5 7 96 4 2 8

output

3 2 1 4

input

5 51 2 1 2 53 1 4 1 5

output

4 2 4 2 5

题意：求b数组中每个元素在a数组中比其小或者等于的数的个数。

sort排序之后，暴力一发超时，我真是菜啊，居然第一反应是暴力。

很简单的二分查找，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，代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[200010],b[200010];int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;++i)scanf("%d",&a[i]);sort(a,a+n);for(j=0;j<m;++j)scanf("%d",&b[j]);for(i=0;i<m;++i){if(i==m-1)printf("%d\n",upper_bound(a,a+n,b[i])-a);else printf("%d ",upper_bound(a,a+n,b[i])-a);} }return 0;}