POJ 1833 permutation (STL)

Arranged

http://poj.org/problem?id=1833

Time limit:1000ms

Memory limit:30000k

Description

Topic Description:

As you know, given a positive integer n, the number of n 1 to n can constitute n! , listing these permutations in a small to large order (dictionary order), such as n=3, listing 1 2 3,1 3 2,2 1 3,2 3 1,3 1 2,3 2 16 permutations.

Task Description:

Give an arrangement, find the arrangement of the following K, and if the last arrangement is encountered, the next 1 rows are ranked 1th, that is, 1 2 3...n.

For example: n = 3,k=2 gives an arrangement of 2 3 1, its bottom 1 arranges 3 1 2, and the next 2 is ranked 3 2 1, so the answer is 3 2 1.

Input

The first line is a positive integer m, represents the number of test data, the following is the M group of test data, the first row of each set of test data is 2 positive integers n (1 <= N < 1024) and K (1<=k<=64), the second row has n positive integers, is 1,2 ... an arrangement of N.

Output

For each group of input data, output a row, n number, separated by a space in the middle, representing the input arrangement of the next K arrangement.

Sample Input

3

3 1

2 3 1

3 1

3 2 1

10 2

1 2 3 4 5 6 7 8 9 10

Sample Output

3 1 2

1 2 3

1 2 3 4 5 6 7 9 8 10

Complete code:

/*469ms,168kb*/
    
#include <cstdio>  
#include <algorithm>  
using namespace std;  
    
int num[1024];  
    
int main (void)  
{  
    int m, n, K;  
    scanf ("%d", &m);  
    while (m--)  
    {  
        scanf ("%d%d", &n, &k);  
        for (int i = 0; i < n; ++i)  
            scanf ("%d", &num[i]);  
        while (k--)  
            next_permutation (num, num + N);  
        printf ("%d", num[0]);  
        for (int i = 1; i < n; ++i)  
            printf ("%d", Num[i]);  
        Putchar (' \ n ');  
    }  
    return 0;  
}

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