hdoj 5500 Reorder the Books 【规律题】
2015-11-14 22:42
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Reorder the Books
[b]Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 924 Accepted Submission(s): 506
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
Source
BestCoder Round #59 (div.1)
[/b]
思路:
就是从后(最大的那个数开始)往前找,看看有多少个相连(如果不相连的话,最终依然得把它重新拿到顶上)的递减的数,如果找到之后用总的数的个数减去递减的个数就是要求的个数了!
代码如下:
#include <stdio.h> #include <string.h> int a[25]; int n; int main() { int T; scanf("%d",&T); while(T--) { scanf("%d",&n); int maxn=0; int t; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); if(a[i]>maxn) { maxn=a[i]; t=i; } } int cnt=1; for(int i=t;i>=1;i--) { if(a[i]==a[t]-1) { cnt++; t=i; } } printf("%d\n",n-cnt); } return 0; }
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