康托展开与逆康托展开算法模板

const int N =21;
ll fac[N];
void init(){
    fac[0] = 1;
    for (int i =1; i < 21; i++) {
        fac[i] = fac[i - 1] * i;
    }
}
ll contor(int n,int s[]) {
    ll rnk = 0, cnt;
    for (int i = 0; i < n; i++) {
        cnt = 0;
        for (int j = i + 1; j < n; j++){
            if (s[i] > s[j])cnt++;
        }
        rnk += cnt * fac[n - i - 1];
    }
    return rnk + 1;//如果排名从1开始
}
//n为数列元素个数，rnk为排名，s为计算出的排列
void revcontor(int n, ll rnk, int s[]) {
    rnk--;
    bool vis[21] = {0};
    ll t; int j;
    for (int i = 0; i < n; i++) {
        t=rnk/fac[n-i-1];
        for (j = 1; j <=n; j++) {
            if (!vis[j]) {
                if (t == 0)break;
                t--;
            }
        }
        s[i] =j;
        vis[j] = 1;
        rnk %= fac[n - i - 1];
    }
}

康托展开与逆康托展开算法模板

原文：https://www.cnblogs.com/gzr2018/p/11301865.html