zoukankan      html  css  js  c++  java
  • Reorder the Books(规律)

    Reorder the Books

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 179    Accepted Submission(s): 125


    Problem Description
    dxy has a collection of a series of books called "The Stories of SDOI",There are n(n19) 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(T30) 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

     分析:每次取最大编号的书籍放在最上面,令ans=最大编号,从下往上找,如果当前编号等于最大编号,就不用抽了,编号数减一,最终编号的值就是最小步数;

    代码:

     1 #include<stdio.h>
     2 int main(){
     3     int T,N,ans,m[20];
     4     scanf("%d",&T);
     5     while(T--){
     6         scanf("%d",&N);
     7         for(int i=1;i<=N;i++)scanf("%d",m+i);
     8         ans=N;
     9         for(int i=N;i;i--){
    10             if(ans==m[i])ans--;
    11         }
    12         printf("%d
    ",ans);
    13     }
    14     return 0;
    15 }
  • 相关阅读:
    [docker]Kubernetes的yaml文件
    [redis]redis-cluster的使用
    [redis]redis-cluster搭建
    [docker]本地仓库的创建的使用
    1W字看懂互联网知识经济
    PHP基础陷阱题(变量赋值)
    PHP不用第三变量交换2个变量的值的解决方法
    PHP中的排序函数sort、asort、rsort、krsort、ksort区别分析
    PHP实现四种基本排序算法
    WEB安全之Token浅谈
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4868811.html
Copyright © 2011-2022 走看看