hdu 5122 K.Bro Sorting (水题）

Problem Description

Matt’s friend K.Bro is an ACMer.

Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.

Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.

There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it. For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N .

 

 

Input

The first line contains only one integer T (T ≤ 200), which indicates the number of test cases. For each test case, the first line contains an integer N (1 ≤ N ≤ 10 ⁶).

The second line contains N integers a _i (1 ≤ a _i ≤ N ), denoting the sequence K.Bro gives you.

The sum of N in all test cases would not exceed 3 × 10 ⁶.

 

 

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of rounds needed to sort the sequence.

 

 

Sample Input

2

5

5 4 3 2 1

5

5 1 2 3 4

 

 

Sample Output

Case #1: 4

Case #2: 1

 

   题意：给一个数列，从小到大按照以下规则排序。从数列中选择一个数向右移直到右边的数比它大就停下。问最少要多少步。

   思路：对于一个数如果它右边有比它小的数，那么它就一定要移动。如果每次移最大的数，那么一个数最多只要移动一次，所以题就简化成数列中有多少个数有右边比它小的数。

           所以就从右边开始循环，没次记录最小值，并用最小值和下一个数比较，如果下一个值比最小值大，结果就加1.

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[1000006];
int main()
{
    int mn,i,t,n,ans,k=1;
    scanf("%d",&t);
    while (t--)
    {
        ans=0;
        scanf("%d",&n);
        for (i=1;i<=n;i++) scanf("%d",&a[i]);
        mn=a[n];
        for (i=n-1;i>=1;i--)
        {
            if (mn>a[i]) mn=a[i];
            else ans++;
        }
        printf("Case #%d: %d\n",k,ans);
        k++;
    }
}