hdoj 5199 Gunner map
2024-10-15 17:45:21
Gunner
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5199
Description
Long long ago, there is a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i−th bird stands on the top of the i−th tree. The trees stand in straight line from left to the right. Every tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know how many birds fall during each shot.
Jack will shot many times, he wants to know how many birds fall during each shot.
a bullet can hit many birds, as long as they stand on the top of the tree with height of H.
Input
There are multiple test cases (about 5), every case gives n,m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times.
In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.
In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.
Please process to the end of file.
[Technical Specification]
1≤n,m≤1000000(106)
1≤h[i],q[i]≤1000000000(109)
All inputs are integers.
Output
For each q[i], output an integer in a single line indicates the number of birds Jack shot down.
Sample Input
4 3 1 2 3 4 1 1 4
Sample Output
1 0 1
HINT
题意
给你n个数m次询问,问你大小为A的数有多少个,第二次问的时候,就直接输出0就好
题解:
map可过,非常轻松(雾
~\(≧▽≦)/~啦啦啦
代码:
#include<cstdio>
#include<map>
using namespace std;
map<int,int> g;
int n,m;
inline int read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
g.clear();
for(int i=;i<n;i++)
{
int x=read();
g[x]++;
}
for(int i=;i<m;i++)
{
int x=read();
if(g.find(x)==g.end())
{
printf("0\n");
}
else
{
printf("%d\n",g[x]);
g[x]=;
}
}
}
}
最新文章
- solar system by HTML5
- MSSQLLocalDB 连接字符串 vs2015
- 如何在 CentOS 7 用 cPanel 配置 Nginx 反向代理
- iOS开发UI篇—无限轮播(功能完善)
- 【Android】线程池原理及Java简单实现
- mysql 将时间戳直接转换成日期时间
- C# vba 操作 Word
- 【EPplus】Column width discrepancy
- OpenFileDialog组件打开文件....待续
- &;#181;C/OS-II版本升级指南
- 移动WEB开发资源
- ASP.NET Core Web APi获取原始请求内容
- 使用Freemarker 实现JSP页面的静态化
- Centos7编译opencv3.4.1
- ajax全局事件
- OI生涯回忆录 2017.9.10~2018.11.11
- <;Scala>;<;For beginners>;
- 201621123008 《Java程序设计》第13周学习总结
- 【windows核心编程】HideProcess
- Combination Sum leetcode java