Educational Codeforces Round 2 B. Queries about less or equal elements 水题

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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原文链接：http://www.cnblogs.com/qscqesze/p/5005382.html

本文解析了CodeForces上的一道题目B，题目要求对于给定的两个整数数组，找出第二个数组中的每个元素，在第一个数组中有多少个小于等于它的元素。通过将第一个数组存入vector并排序，再使用二分查找的方法来解决该问题。

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B. Queries about less or equal elements

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/600/problem/B

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 athat 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

5 4
1 3 5 7 9
6 4 2 8

Sample Output

3 2 1 4

HINT

题意

给你一个数组，然后有Q次询问，问你在数组中，有多少个数比他小

题解：

把数组扔进一个vector里面，排序，然后二分就好了~

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define maxn 200005
int a[maxn];
int b[maxn];
vector<int> Q;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        Q.push_back(a[i]);
    }
    sort(Q.begin(),Q.end());
    for(int i=1;i<=m;i++)
    {
        int p;scanf("%d",&p);
        int x = upper_bound(Q.begin(),Q.end(),p)-Q.begin();
        printf("%d\n",x);
    }
}