Expand kangtuo and expand kangtuo

Expand kangtuo and expand kangtuo

1. Explanations of conto

Kanto expansion is a special hash function.

Expand an integer x into the following form:

X = a [n] * n! + A [n-1] * (n-1 )! +... + A [2] * 2! + A [1] * 1!

Where, a is an integer and 0 <= a <I, I = 1, 2,..., n

{1, 2, 3, 4,..., n} indicates the arrangement of 1, 2, 3,..., n, for example, {1, 2, 3}. A total of 6 entries are arranged in ascending order. 123 132 213 231 312 321.

The number 1, 2, 4, 5, and 6 correspond to a given number.

The relationships between them can be found in Kanto.

If you want to know the number of the largest numbers in 321 is {1, 2, 3}, you can consider this as follows:

The first digit is 3. When the first digit is less than 3, the number is smaller than 321, for example, 123 and 213. The number smaller than 3 is 1 and 2. So there are 2*2! . Let's look at the number smaller than the second digit 2: only one number smaller than 2 is 1, so there is 1*1! = 1 so the number of {1, 2, 3} arrays smaller than 321 is 2*2! + 1*1! = 5
. So 321 is a large number of 6th. 2*2! + 1*1! It is Kanto expansion.

For another example, 1324 is the largest number in the {1, 2, 3, 4} number. The first is the number where 1 is smaller than 1. The number is 0 0*3! The second digit is 3 less than 3. The numbers are 1 and 2, but 1 is already in the first place, so there is only one digit 2 1*2! . The third digit is 2 and the number less than 2 is 1, but 1 is in the first place, so

There are 0 numbers 0*1! So there are 0*3 smaller than 1324! + 1*2! + 0*1! = 2, 1324 is the third large number.

(1) identify the number of 96th

First, use 96-1 to get 95

Remove 4 with 95! Get 3 + 23

Remove 3 with 23! Get more than 3 5

Remove 2 with 5! Get 2 more than 1

Remove 1 with 1! The result is that there are three numbers in the range of 1 to 0, and the smaller the number is 4.

So the first is 4.

There are three numbers that are smaller than 4, but 4 has already appeared before, so it is 5 (because 4 has appeared before, so the actual number smaller than 5 is 3)

There are two smaller numbers than it: 3

The number of one smaller than it is 2.

The last number can only be 1

So the number is 45321

C ++ code:

/* Input1 51 2 3 51 51 2 3 5 42 511 54 5 3 2 12 595 OutputCantor = 0 Cantor = 1 Series: 1 2 3 5 4 Cantor = 95 series: 4 5 3 2 1 */# include <iostream> # include <cstdio> # include <string> # include <cmath> # include <queue> # include <stack> # include <map> # include <cstring> # include <algorithm> # define rep (I, a, B) for (int I = (a); I <(B); I ++) # define rev (I, a, B) for (int I = (a); I >= (B); I --) # define clr (a, x) memset (a, x, sizeof a) # define inf 0x3f3f3f3f Typedef long LL; using namespace std; const int mod = 1e9 + 7; const int maxn = 2005; const int maxm = 4005; int fac [] =, 24,120,720,504, 362880,}; int cantor (int * s, int n) {int num = 0; for (int I = 0; I <n-1; I ++) {int tmp = 0; for (int j = I + 1; j <n; j ++) if (s [j] <s [I]) tmp ++; num + = fac [n-i-1] * tmp;} return num;} void _ cantor (int * s, int n, int x) {bool tmp [n + 1]; clr (tmp, 0); for (int I = n-1; I> = 0; I --) {int k = x/fac [I]; x % = fac [I]; int j = 1; for (int sum = 0; sum <k | tmp [j]; j ++) if (! Tmp [j]) sum + +; s [n-1-i] = j; tmp [j] = 1 ;}} int main () {int flag, n; while (~ Scanf ("% d", & flag, & n) {if (flag = 1) {int s [n]; for (int I = 0; I <n; I ++) scanf ("% d", & s [I]); printf ("Cantor = % d \ n", cantor (s, n);} else if (flag = 2) {int x, ans [n]; scanf ("% d", & x); _ cantor (ans, n, x); cout <"sequence:"; for (int I = 0; I <n; I ++) printf ("% d % c ", ans [I], I = n-1? '\ N': '') ;}} return 0 ;}

Reference: http://www.cnblogs.com/hxsyl/archive/2012/04/11/2443009.html