Educational Codeforces Round 2 B. Queries about less or equal elements 水题
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=;i<=n;i++)
{
scanf("%d",&a[i]);
Q.push_back(a[i]);
}
sort(Q.begin(),Q.end());
for(int i=;i<=m;i++)
{
int p;scanf("%d",&p);
int x = upper_bound(Q.begin(),Q.end(),p)-Q.begin();
printf("%d\n",x);
}
}
Educational Codeforces Round 2 B. Queries about less or equal elements 水题的更多相关文章
- Educational Codeforces Round 2 B. Queries about less or equal elements
打开题目连接 题意:给2个数组(无序的)啊a,b,判断b数组中的每一个元素大于a数组中个数. ACcode: #include <iostream> #include <vector ...
- Educational Codeforces Round 76 (Rated for Div. 2) C. Dominated Subarray 水题
C. Dominated Subarray Let's call an array
- Educational Codeforces Round 76 (Rated for Div. 2) B. Magic Stick 水题
B. Magic Stick Recently Petya walked in the forest and found a magic stick. Since Petya really likes ...
- Educational Codeforces Round 69 (Rated for Div. 2) C. Array Splitting 水题
C. Array Splitting You are given a sorted array
- Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题
Educational Codeforces Round 71 (Rated for Div. 2)-E. XOR Guessing-交互题 [Problem Description] 总共两次询 ...
- Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)
Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...
- Codeforces Round #373 (Div. 2) C. Efim and Strange Grade 水题
C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just re ...
- Codeforces Round #185 (Div. 2) A. Whose sentence is it? 水题
A. Whose sentence is it? Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/ ...
- Codeforces Round #373 (Div. 2) A. Vitya in the Countryside 水题
A. Vitya in the Countryside 题目连接: http://codeforces.com/contest/719/problem/A Description Every summ ...
随机推荐
- TCP/IP详解学习笔记(4)-ICMP协议,ping和Traceroute
1.IMCP协议介绍 前面讲到了,IP协议并不是一个可靠的协议,它不保证数据被送达,那么,自然的,保证数据送达的工作应该由其他的模块来完成.其中一个重要的模块就是ICMP(网络控制报文)协议. 当传送 ...
- 云计算服务模型,第 3 部分: 软件即服务(PaaS)
英文原文:Cloud computing service models, Part 3: Software as a Service 软件即服务 (SaaS) 为商用软件提供基于网络的访问.您有可能已 ...
- ios实现类似魔兽小地图功能 在
写了一个类似魔兽小地图功能的控件. 比如你有一个可以放大缩小的scrollView.会在里面进行一些放大缩小,点击里面的按钮呀,等操作. 这个小地图控件.就会和你的大scrollView同步.并有缩略 ...
- POJ 1860 Currency Exchange
题意:有n种货币,可以互相兑换,有m个兑换规则,兑换规则给出汇率r和手续费c,公式为b = (a - c) * r,从a货币兑换为b货币,问能不能通过不断的兑换赚钱,兑换期间手中的钱数不可以为负. 解 ...
- HDU 5429 Geometric Progression
题意:给出一个大数数列,问是不是等比数列. 解法:拿java大数搞,注意全是0的情况也是Yes.我把公比用分数表示了,灰常麻烦,题解说只要判a[i - 1] * a[i + 1] == a[i] * ...
- Oracle :一次数据库连接,返回多个结果集
1. 一次数据库连接,返回多个结果集 1.1 建立包规范 create or replace package QX_GDJTJ is -- Author : xxx -- Created : 2012 ...
- C# 中LinkLabel的简单使用
界面中加入一个LinkLabel控件
- Android Scrollview 内部组件android:layout_height="fill_parent"无效的解决办法
Found the solution myself in the end. The problem was not with the LinearLayout, but with the Scrol ...
- C#网络资源列表
1, http://club.topsage.com/thread-371996-1-1.html
- Cake slicing
题意: n*m的方格中有k个点,现在要把方格分开使得每个点在一个部分,每分一次花费边长的费用,求完成花的最小费用 分析: dp[sx][sy][ex][ey]表示分割起点(sx,sy)终点(ex,ey ...