题目分析：首先数列长度为n,且为1~n中不重复的数字，也就意味着每个数字必须出现且只出现一次。首先，记录下输入中已经出现过的数字，再找出还未出现的数字，通过对未出现的数字进行全排列，再插入到原数组中，将原数组中的0替代掉，然后对替代后的数组进行判断，暴力计算合格排列为K的数列个数。代码如下

#include <iostream>
#include <algorithm> 
#include <vector>

const int N = 100;
int vis[N];
using namespace std;

int isOk(int *num,int *queshi, int n,int k)
{
    int tmp[N], acc = 0;
    for (int i = 1, j = 1; i <= n; i++) //构建新数组，将原数组中的0替代
    {
        if (num[i] == 0)
        {
            tmp[i] = queshi[j++];
        }
        else{
            tmp[i] = num[i];
        }

        //printf("%d ",tmp[i]);
    }
    for (int i = 1; i<=n - 1; i++)   //对新数组进行暴力判断合法排列数目
    {
        for (int j = i + 1; j<=n; j++)
        {
            if (tmp[i]<tmp[j])
                acc++;
        }
    }
    if (acc == k)
        return 1;
    else
        return 0;
}
void permutation(int *queshi, int *num, int n, int len, int index, int k, int &count )
{ 
    //对数组queshi全排列
        if (index > len){//全排列结束
            for (int i = 1; i <= len; i++)
            {
                cout << queshi[i];
            }
            cout << endl;

            if (isOk(num, queshi,n,k) == 1)
            {
                count++;
            }

        }
        else
            for (int i = index; i<=len; ++i){
            //将第i个元素交换至当前index下标处
            swap(queshi[index], queshi[i]);

            //以递归的方式对剩下元素进行全排列
            permutation(queshi, num, n, len, index + 1, k,count);

            //将第i个元素交换回原处
            swap(queshi[index], queshi[i]);
            }
    }


    int main()
    {
        int n, k;
        cin >> n >> k;
        int temp;
        int count = 0;
        int num[N];

        int zero = 0;
        for (int i = 0; i <= n; i++){
            vis[i] = 0;
        }

        for (int i = 1; i <= n; i++)
        {
            cin >> temp;
            num[i] = temp;
            vis[temp] = 1;          //记录已经出现的数字
            if (temp == 0)
                zero++;             //记录0的个数，同时也是数组queshi的范围1~zero
        }
        int queshi[N];
        int v = 1;
        for (int i = 1; i <= n; i++)
        {
            if (vis[i] == 0)
            {
                queshi[v++] = i; //记录没有出现的数字
            }

        }
        ///cout << zero;

        permutation(queshi,num, n, zero,1, k,count);
        cout << count << endl;
        //cout << count << endl;
        system("pause");

}