K.Bro Sorting

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 571 Accepted Submission(s): 300

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 ≤ 106).

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

The sum of N in all test cases would not exceed 3 × 106.

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

Hint

In the second sample, we choose “5” so that after the first round, sequence becomes “1 2 3 4 5”, and the algorithm completes.

题意

题意就是告诉你，某个人发明了一种新的排序算法，就是在这个序列中，随便选一个数，然后与后面与他相邻的数进行比较，如果大于后面的，就交换，直到不能交换为止
然后问你，这种操作最少需要多少次

题解

我们从后面开始找，假设最小值是最后一个数，然后让他与前面的比较，如果前面的数比他小的话，就更新最小值，否则就ans++
至于为什么，我们可以很容易证明，已经交换过的后面的序列，一定是从小到大排好了的，所以这样搞是可行的

吐槽

hdu用G++交的话，读入会很慢，然后T掉

代码

int a[maxn];
int main()
{
    int t;
    RD(t);
    REP_1(ti,t)
    {
        int n;
        RD(n);
        REP(i,n)
            RD(a[i]);
        int ans=0;
        int minn=a[n-1];
        for(int i=n-2;i>=0;i--)
        {
            if(minn<a[i])
                ans++;
            else
                minn=a[i];
        }
        printf("Case #%d: %d\n",ti,ans);
    }
}