Sequence in the Pocket

最新推荐文章于 2024-05-29 15:55:53 发布

原创最新推荐文章于 2024-05-29 15:55:53 发布 · 475 阅读

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本文探讨了一种算法，旨在通过最少的操作次数将任意整数序列调整为非递减顺序。具体而言，操作涉及选择序列中的任意元素并将其移动到序列的开头。文章提供了示例输入和输出，详细解释了算法的实现过程，并附带了完整的C++代码实现。

Time Limit: 1 Second Memory Limit: 65536 KB

DreamGrid has just found an integer sequence in his right pocket. As DreamGrid is bored, he decides to play with the sequence. He can perform the following operation any number of times (including zero time): select an element and move it to the beginning of the sequence.

What's the minimum number of operations needed to make the sequence non-decreasing?

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains an integer (), indicating the length of the sequence.

The second line contains integers (), indicating the given sequence.

It's guaranteed that the sum of of all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the answer.

Sample Input

Sample Output

2
0

Hint

For the first sample test case, move the 3rd element to the front (so the sequence become {2, 1, 3, 4}), then move the 2nd element to the front (so the sequence become {1, 2, 3, 4}). Now the sequence is non-decreasing.

For the second sample test case, as the sequence is already sorted, no operation is needed.

用数组b保存理想顺序（升序），由大到小依次找有多少是满足顺序的，这部分不用动，其余的都要移。可手动验证。

#include<iostream>
#include<algorithm>
const int maxn = 1e5+2;
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--){
        int a[maxn],b[maxn];
        int n;
        cin>>n;
        for(int i = 0;i < n;i++){
            cin>>a[i];
            b[i] = a[i];
        }
        sort(b,b+n);
        int l = n-1;
        int num = 0;
        for(int i = n-1;i >= 0;i--){
            while(a[l] != b[i]&&l >= 0)
                l--;
            if(a[l] == b[i]){
                num++;
                l--;
            }
            if(l < 0)
                break;
        }
        cout<<n-num<<endl;
    }
}