CodeForces600B(二分查找)

来源：互联网发布：文艺知乎编辑：程序博客网时间：2024/05/22 01:28

Queries about less or equal elements
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Submit

Status

Practice

CodeForces 600B
Description
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.

Sample Input
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
之前想了一个O(n)的算法就是把两个数组排序最后扫一遍就行了，太慢。最后用二分logn过了

#include "cstring"#include "cstdio"#include "map"#include "algorithm"#include "iostream"#include "string.h"#include "cstdlib"using namespace std;int a[200002];int b[200002];int Compare(const void *elem1, const void *elem2){    return *((int *)(elem1)) - *((int *)(elem2));}int main(){    int n,m;    scanf("%d%d",&n,&m);    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        for(int i=0;i<m;i++)        {            scanf("%d",&b[i]);        }        sort(a,a+n);        for(int i=0;i<m;i++)        {            printf("%ld ",upper_bound(a,a+n,b[i])-a);        }    }}

0 0