全排列方法二（康托展开）

#include
const int fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};///阶乘

int KT(int s[], int n)
{
	int i, j, cnt, sum;
	sum = 0;
	for (i = 0; i < n; ++i)
	{
		cnt = 0;
		for (j = i + 1; j < n; ++j)
			if (s[j] < s[i]) ++cnt;
		sum += cnt * fac[n - i - 1];
	}
	return sum;
}

int main()
{
	int a[] = {3, 5, 7, 4, 1, 2, 9, 6, 8};
	printf("%d\n", 1 + KT(a, sizeof(a) / sizeof(*a))); ///1+98884
}

#include
#include
const int fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};///阶乘

bool vis[10];

///n为ans大小，k为全排列的编码
void invKT(int ans[], int n, int k)
{
	int i, j, t;
	memset(vis, 0, sizeof(vis));
	--k;
	for (i = 0; i < n; ++i)
	{
		t = k / fac[n - i - 1];
		for (j = 1; j <= n; j++)
			if (!vis[j])
			{
				if (t == 0) break;
				--t;
			}
		ans[i] = j, vis[j] = true;
		k %= fac[n - i - 1];///余数
	}
}

int main()
{
	int a[10];
	invKT(a, 5, 16);
	for (int i = 0; i < 5; ++i)
		printf("%d ", a[i]);///1 4 3 5 2
}