Reorder the Books

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 919    Accepted Submission(s):
501

Problem Description
dxy has a collection of a series of books called "The
Stories of SDOI",There are n(n≤19)

books in this series.Every book has a number from 1

to n

.

dxy puts these books in a book stack with the order of their numbers
increasing from top to bottom. dxy takes great care of these books and no one is
allowed to touch them.

One day Evensgn visited dxy's home, because dxy
was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn
was curious about this series of books.So he took a look at them. He found out
there was a story about "Little E&Little Q". While losing himself in the
story,he disrupted the order of the books.

Knowing that dxy would be back
soon,Evensgn needed to get the books ordered again.But because the books were
too heavy.The only thing Evensgn could do was to take out a book from the book
stack and and put it at the stack top.

Give you the order of the
disordered books.Could you calculate the minimum steps Evensgn would use to
reorder the books? If you could solve the problem for him,he will give you a
signed book "The Stories of SDOI 9: The Story of Little E" as a gift.

 
Input
There are several testcases.

There is an
positive integer T(T≤30)

in the first line standing for the number of testcases.

For each
testcase, there is an positive integer n

in the first line standing for the number of books in this
series.

Followed n

positive integers separated by space standing for the order of the disordered
books,the ith

integer stands for the ith

book's number(from top to bottom).

Hint:
For the first
testcase:Moving in the order of book3,book2,book1

,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4)

,and this is the best way to reorder the books.
For the second testcase:It's
already ordered so there is no operation needed.

 
Output
For each testcase,output one line for an integer
standing for the minimum steps Evensgn would use to reorder the books.
 
Sample Input
2
4
4 1 2 3
5
1 2 3 4 5
 
Sample Output
3
0
 
题意:一摞书顺序乱了,让你按照序号从小到大的顺序将其排列整齐,一次只能拿一本书,且必须放在最上边,问最少多少步可以实现
题解:先算出已经按顺序排好的书本的个数sum(注意这个序列并不一定连续,而且必须是从某个数到n的序列),用n减去sum即可
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 100100
using namespace std;
int a[MAX];
int main()
{
int n,m,t,i,j;
scanf("%d",&t);
int maxx,ans;
while(t--)
{
scanf("%d",&n);
maxx=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(maxx<a[i])
{
maxx=a[i];
ans=i;
}
}
int ant=ans,sum=0;
for(j=ans;j>=1;j--)
{
if(a[j]==a[ant]-1)
{
sum++;
ant=j;
}
}
printf("%d\n",n-sum-1);
}
return 0;
}

  

hdoj 5500 Reorder the Books的更多相关文章

  1. BestCoder Round 59 (HDOJ 5500) Reorder the Books

    Problem Description dxy has a collection of a series of books called “The Stories of SDOI”,There are ...

  2. hdu 5500 Reorder the Books

    http://acm.hdu.edu.cn/showproblem.php?pid=5500 Reorder the Books Time Limit: 4000/2000 MS (Java/Othe ...

  3. HDU 5500 Reorder the Books 贪心

    Reorder the Books Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php? ...

  4. hdu 5500 Reorder the Books(规律)

    题意:   有一个1→n的排列形成的数列,我们要用最少的操作次数把这个数列从小到大排序,每次操作都是把一个数放到整个数列的最前面. 思路:        首先最大的数n是不用操作的(其他数操作好了,n ...

  5. HDU 5500 Reorder the Books (水题)

    题意: 有n本书,编号为1~n,现在书的顺序乱了,要求恢复成有序的样子,每次只能抽出其中一本并插到最前面,问最少需要多少抽几次? 思路: 如果pos[i]放的不是书i的话,则书i的右边所有的书都必须抽 ...

  6. Reorder the Books -- hdu -- 5500

    http://acm.hdu.edu.cn/showproblem.php?pid=5500 Reorder the Books Time Limit: 4000/2000 MS (Java/Othe ...

  7. Reorder the Books(规律)

    Reorder the Books Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Othe ...

  8. BestCoder Round #59 (div.2) B. Reorder the Books 想法题

    Reorder the Books 问题描述 dxy家收藏了一套书,这套书叫<SDOI故事集>,<SDOI故事集>有n(n\leq 19)n(n≤19)本,每本书有一个编号,从 ...

  9. Reorder the Books-HDU5500

    Problem Description dxy has a collection of a series of books called "The Stories of SDOI" ...

随机推荐

  1. redis 性能监控和排查

    最近项目中接连遇到redis出现瓶颈的问题,现在把排查的一些经验记录下来备查,本篇只是思路的整理,不涉及具体的使用. 大体的思路如下: 1.通过slow log查看 参考 http://www.cnb ...

  2. http://nxlhero.blog.51cto.com/962631/1666250?plg_nld=1&plg_uin=1&plg_auth=1&plg_nld=1&plg_usr=1&plg_vkey=1&plg_dev=1

    http://nxlhero.blog.51cto.com/962631/1666250?plg_nld=1&plg_uin=1&plg_auth=1&plg_nld=1&am ...

  3. Weblogic下部署的应用,当更新文件时需要重新安装部署

    JSP页面检查(秒):-1 Servlet重新加载检查(秒):-1 -1说明从不检查,故当更新文件时,需要重新部署,或重新安装部署.

  4. php中对共享内存,消息队列的操作

    http://www.cnblogs.com/fengwei/archive/2012/09/12/2682646.html php作为脚本程序,通常生命周期都很短,如在web应用中,一次请求就是ph ...

  5. JavaScript代码调试

    怎么在浏览器中调试JavaScript代码呢?首先,你需要安装Google Chrome浏览器,Chrome浏览器对开发者非常友好,可以让你方便地调试JavaScript代码.安装后,随便打开一个网页 ...

  6. Ubuntu 学习笔记

    1.   ubuntu开启root账号,设置分配很简单,只要为root设置一个root密码就行了: $ sudo passwd root 之后会提示要输入root用户的密码,连续输入root密码,再使 ...

  7. Windows平台下的session0创建进程的问题与解决办法

    很多博客都有记载如何在session0下创建进程的办法,也就是使用CreateProcessAsUser.但是这个要求服务的进程有SE_INCREASE_QUOTA_NAME和SE_ASSIGNPRI ...

  8. s3c6410 开发板Linux系统支持 K9GAG08U0E的方法

    由于NandFlash硬件升级比较快,公司去年一直在使用三星的K9GAG08U0D,现在MLC NandFlash 升级到了第二代,K9GAG08U0D 很快就会处在停产的状态,未雨绸缪,公司选型了K ...

  9. MySQL优化器 limit影响的case

    测试的用例中,因为limit的大小不同,而产生了完全不同的执行计划: 1. 测试case: create table t1 ( f1 ) not null, f2 ) not null, f3 ) n ...

  10. 根据block取出页号buf_block_get_page_no

    /*********************************************************************//** Gets the page number of a ...