全排列方法二（康托展开）

#include<cstdio>
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<cstdio>
#include<cstring>
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
}