Gunner II



Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1244    Accepted Submission(s): 486



Problem Description

Long long ago, there was 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 nearest bird which stands in the tree with height H will falls.

Jack will shot many times, he wants to know which bird will fall during each shot.

 

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]

All input items are integers.

1<=n,m<=100000(10^5)

1<=h[i],q[i]<=1000000000(10^9)

 

Output

For each q[i], output an integer in a single line indicates the id of bird Jack shots down. If Jack can’t shot any bird, just output -1.

The id starts from 1.

 

Sample Input

5 5

1 2 3 4 1

1 3 1 4 2

 

Sample Output

1

3

5

4

2

//这题主要思路就是利用一个结构体将高度与序号记录下来
//再利用一个数组将高度排序而且标记高度出现的顺序 #include <stdio.h>
#include <algorithm>
using namespace std;
int flag[100010],h[100010]; struct bird
{
int n,num;
}p[100010]; bool cmp(const bird &a,const bird &b)
{
return a.n!=b.n? a.n<b.n:a.num<b.num;
} int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;i++ )
{
scanf("%d",&p[i].n);
p[i].num=i+1;
}
sort(p,p+n,cmp);
for(int i=0;i<n;i++)
{
flag[i]=i; //此时的flag数组是标记高度第一次出现的位置
h[i]=p[i].n;
}
while(m--)
{
int x;
scanf("%d",&x);
int q=lower_bound(h,h+n,x)-h; //查询到的也是第一次出现的次数
if(h[flag[q]]!=x)
{
printf("-1\n");
continue;
}
else
printf("%d\n",p[flag[q]].num);
flag[q]++; //因为查询过一次了 所以递增到下一个高度
}
}
return 0;
}

HDU5233的更多相关文章

  1. hdu5233 Gunner II

    Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting ver ...

  2. Beatcoder#39+#41+#42

    HDU5211 思路: 倒着更新每个数的约数,更新完要把自己加上,以及1的情况? //#include <bits/stdc++.h> #include<iostream> # ...

随机推荐

  1. hdu 4997 Biconnected

    这题主要是计算连通子图的个数(c)和不连通子图的个数(dc)还有连通度为1的子图的个数(c1)和连通度为2以上的子图的个数(c2)之间的转化关系 主要思路大概例如以下: 用状态压缩的方法算出状态为x的 ...

  2. iframe是否缓存页面探究

    近期手里有个项目须要用iframe来调用每天都会变化的页面,后来想到iframe会不会缓存页面呢.于是写了个demo论证了下,结果例如以下: iframe的src假设是静态页面,就有可能会缓存.由于静 ...

  3. OpenST Basic tool library

    /***************************************************************************** * OpenST Basic tool l ...

  4. Atom介绍和安装步骤

    Atom是全然基于web技术开发而成的一款编辑器,其底层架构依赖于chromium,google chrome浏览器也是基于此.编辑器的每一个窗体都是本地渲染的web页面,而且其风格与时下流行的sub ...

  5. [SCOI 2009] 生日快乐

    [题目链接] https://www.lydsy.com/JudgeOnline/problem.php?id=1024 [算法] 直接DFS,即可 [代码] #include<bits/std ...

  6. Java-MyBatis:MyBatis XML 文件

    ylbtech-Java-MyBatis:MyBatis XML 文件 1.返回顶部 1. Mapper XML 文件 MyBatis 的真正强大在于它的映射语句,也是它的魔力所在.由于它的异常强大, ...

  7. 7.matlab字符串分析

    1 字符串处理函数 clc; clear all; str='My name is Robin.'; disp(str); %字符串的输出 str_size=size(str) %字符串的长度 str ...

  8. Spark SQL概念学习系列之性能调优

    不多说,直接上干货! 性能调优 Caching Data In Memory Spark SQL可以通过调用sqlContext.cacheTable("tableName") 或 ...

  9. SQL Server 获取两个日期间的日期

    declare @start datetime declare @end datetime set @start = '2018-01-25' set @end = '2018-02-03' sele ...

  10. Core篇——初探Core的认证,授权机制

    目录 1.Cookie-based认证的实现 2.Jwt Token 的认证与授权 3.Identity Authentication + EF 的认证 Cookie-based认证的实现 cooki ...