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/*******************************************************************************
Finding permutation by its length and number in lexicographical order
Works in O(n^2)
Based on problem 190 from informatics.mccme.ru
http://informatics.mccme.ru/moodle/mod/statements/view.php?chapterid=190#1
*******************************************************************************/
#include <iostream>
#include <fstream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cassert>
#include <utility>
#include <iomanip>
using namespace std;
const int MAXN = 20;
int n, num;
bool used[MAXN];
int f[MAXN];
void permutationByNumber(int n, int num) {
for (int i = 1; i <= n; i++)
used[i] = false;
int cur = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (used[j])
continue;
int add = f[n - i];
if (cur + add >= num) {
used[j] = true;
printf("%d ", j);
break;
}
cur += add;
}
}
printf("\n");
}
int main() {
//assert(freopen("input.txt","r",stdin));
//assert(freopen("output.txt","w",stdout));
scanf("%d", &n);
scanf("%d", &num);
f[0] = 1;
for (int i = 1; i <= n; i++) {
f[i] = f[i - 1] * i;
}
permutationByNumber(n, num);
return 0;
}