Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 21268 | Accepted: 8049 |
Description
任务描述:
给出某个排列,求出这个排列的下k个排列,如果遇到最后一个排列,则下1排列为第1个排列,即排列1 2 3…n。
比如:n = 3,k=2 给出排列2 3 1,则它的下1个排列为3 1 2,下2个排列为3 2 1,因此答案为3 2 1。
Input
Output
Sample Input
33 12 3 13 13 2 110 2 1 2 3 4 5 6 7 8 9 10
Sample Output
3 1 21 2 31 2 3 4 5 6 7 9 8 10
Source
#include <map>#include <set>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <vector>#include <string>#include <cstring>#include <iterator>#include <iostream>#include <algorithm>#define debug(a) cout << #a << " " << a << endlusing namespace std;const int maxn = 2*1e4 + 10;const int mod = 10000;typedef long long ll;ll num[maxn];int main() { std::ios::sync_with_stdio(false); ll t; scanf("%lld",&t); while( t -- ) { ll n, m; scanf("%lld %lld", &n, &m); for( ll i = 0; i < n; i ++ ) { scanf("%lld",&num[i]); } for( ll i = 0; i < m; i ++ ) { next_permutation( num, num + n ); } copy( num, num+n-1, ostream_iterator<ll>(cout, " ")); cout << num[n-1] << endl; } return 0;}