CodeForces - 600B Queries about less or equal elements

CodeForces - 600B Queries about less or equal elements
题目描述：
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.

Examples
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

题目大意：一个一个处理数组b中的数，在数列a中找到小于或者等于数组b的数的个数。
思路：其实就是upper_bound()

#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>

using namespace std;
int n , m;
int a[200010];
int b[200010];
int c[200010];
int main(){
    while(~scanf("%d%d",&n,&m)){
        for(int i = 0 ; i < n ; ++i)
            scanf("%d",&a[i]);
        sort(a , a+n);
        for(int i = 0 ; i < m ; ++i){
            scanf("%d",&b[i]);
            c[i] = upper_bound(a,a+n,b[i]) - a;
        }
        for(int i = 0 ; i < m - 1 ; ++i){
            printf("%d ",c[i]);
        }
        printf("%d\n",c[m-1]);
    }
    return 0;
}