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
Hint
In the second sample, we choose “5” so that after the ?rst round, sequence becomes “1 2 3 4 5”, and the algorithm completes. source：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=94498#problem/K这道题是从后往前做的，我想从前往后，试了确实不行，但是没想通为什么不行网上的比较好的解释： 1：给出一个数列，每次随机选择一个数，按照冒泡排序的方法去交换，问这种排序最快需要几个回合 思路：如果某个数右边有比他小的数字，那这个数一定要被扔到后边去。所以线性方法统计有多少个这样的数字即可
2：从后往前，存在一组递增子数列则ans加1，因为对于每个这样的递增子序列我们都最少需要选择一次将其排序
/*
问题：给定一个序列，用给定的排序方法把它从小到大排列
给定方法：任意选择一个数，依次找后边比他小的数，并交换这两个数，直到后边没哟比他小的数
称这样的一次操作为一轮
问多少轮操作之后排序成功
分析：从后往前找递增序列数，对于每个递增序列，至少需要移动一次
*/
#include <iostream>
#include <stdio.h>
#define max_num 1000000+10
int num[max_num];
int main()
{
int t, case_num = 0;
scanf("%d", &t);
while(t--)
{
int n, cnt = 0, mi;
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%d", &num[i]);
mi = num[n-1];
for(int i = n-2; i >= 0; i--)
{
if(num[i] < mi)
mi = num[i];
else
cnt++;
}
printf("Case #%d: %d\n", ++case_num, cnt);
}
return 0;
}