permutation 1

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题目链接

题意

找出一个n的全排列，让其后向差分序列字典序最小。

思路

1≤K≤min(10000,n!)，注意到这一点，我们可以知道当n大于8的时候，我们前面几项应该是n,1,2,3,4,…直到最后八项。
最后八项怎么确定呢，看到这个k范围很小，不妨枚举出所有全排列，然后排个序去找第k大。

Code

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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...)                           \
    do                                      \
    {                                       \
        cout << "\033[33;1m" << #x << "->"; \
        err(x);                             \
    } while (0)
void err()
{
    cout << "\033[39;0m" << endl;
}
template <template <typename...> class T, typename t, typename... A>
void err(T<t> a, A... x)
{
    for (auto v : a) cout << v << ' ';
    err(x...);
}
template <typename T, typename... A>
void err(T a, A... x)
{
    cout << a << ' ';
    err(x...);
}
#else
#define dbg(...)
#endif

const int N = 25;
const int maxn = 100005;
int arr[maxn][25];
int cha[maxn][25];
int id[maxn];
int fac[10];
int a[N];
bool cmp1(int x, int y)
{
    for (int i = 1;; i++)
    {
        if (cha[x][i] < cha[y][i])
            return true;
        if (cha[x][i] > cha[y][i])
            return false;
    }
    return false;
}

int main()
{
    int T;
    fac[0] = 1;
    for (int i = 1; i <= 8; i++)
        fac[i] = fac[i - 1] * i;
    scanf("%d", &T);
    while (T--)
    {
        int n, k;
        scanf("%d%d", &n, &k);
        int state = 0;
        int res = n;
        int pre = -1;
        if (n > 8)
        {
            printf("%d ", n);
            state |= 1 << (n - 1), pre = n;
            for (int i = 1; n - i - 1 >= 8; i++)
                printf("%d ", i), state |= 1 << (i - 1), pre = i;
            res = 8;
        }
        int cur = 0;
        for (int i = 0; i < n; i++)
        {
            if (!((1 << i) & state))
                a[cur++] = i + 1;
        }
        for (int i = 1; i <= fac[res]; i++)
        {
            id[i] = i;
        }
        int t = 0;
        if (pre == -1)
        {
            do
            {
                t++;
                for (int i = 0; i < cur; i++)
                    arr[t][i] = a[i];
            } while (next_permutation(a, a + cur));
            for (int i = 1; i <= fac[res]; i++)
            {
                for (int j = 1; j < res; j++)
                    cha[i][j] = arr[i][j] - arr[i][j - 1];
            }
            sort(id + 1, id + fac[res] + 1, cmp1);
            /* puts("askjdal");
            for (int i = 1; i <= fac[res]; i++)
            {
                for (int j = 0; j < res; j++)
                    printf("%d ", arr[id[i]][j]);
                putchar('\n');
            }
            //puts("ajksdfhajkhdfj");*/
            for (int i = 1; i <= res; i++)
                printf("%d%c", arr[id[k]][i - 1], i == res ? '\n' : ' ');
        }
        else
        {
            do
            {
                t++;
                arr[t][0] = pre;
                for (int i = 0; i < cur; i++)
                    arr[t][i + 1] = a[i];
            } while (next_permutation(a, a + cur));
            for (int i = 1; i <= fac[res]; i++)
            {
                for (int j = 1; j <= res; j++)
                    cha[i][j] = arr[i][j] - arr[i][j - 1];
            }
            sort(id + 1, id + fac[res] + 1, cmp1);
            for (int i = 1; i <= res; i++)
                printf("%d%c", arr[id[k]][i], i == res ? '\n' : ' ');
        }
    }
    return 0;
}
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