CSU 1563:lexicography (mathematical counting problem)

1563:lexicography time limit: 1 Sec Memory Limit: MB
Submit: 342 Solved: 111
[Submit] [Status] [Web Board] Description

An anagram of a string was any string that can be formed using the same letters as the original. (We consider the original string an anagram of itself as well.) For example, the string ACM have the following 6 anagrams, as given in alphabetical order:

Acm
Amc
CAM
Cma
Mac
Mca
As another example, the string ICPC has the and the following anagrams (in alphabetical order):

CCIP
Ccpi
Cicp
CIPC
CPCI
CPIC
ICCP
ICPC
IPCC
PCCI
Pcic
Picc
Given a string and a rank K, you is to determine the Kth such anagram according to alphabetical order.

Input

Each test case is designated on a single line containing the original word followed by the desired rank K. Words would Use uppercase letters (i.e., A through Z) and would has length at the most 16. The value of K would be in the range from 1 to the number of distinct anagrams of the given word. A Line of the form "# 0" designates the end of the input.

Output

For each test, display the Kth anagram of the original string.

Sample Input

ACM 5ICPC 12REGION 274# 0

Sample Output

Macpiccignore

HINT

The value of K could is almost 245 in the largest tests, so you should use type Long in Java, or type long long in C + + to Store K.

Permutation combination count problem, n number is all arranged as n!, if X has T, then the whole arrangement is n!/(t!) Then enumerate the current bit is the case of I, and then always minus the case of I, if the case of <i,//Then this one is I #include<iostream> #include <cstdio> #include < cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <vector> #include <set> #include <map> #define L (x) (x<<1) #define R (x) (x<<1|1) #define MID (x, Y)  ((x+y) >>1) #define EPS 1e-8typedef long long ll; #define FRE (i,a,b) for (i = A; I <b; i++) #define FREE (i,b,a) for (i = b; I >= a;i--) #define MEM (T, v) memset ((t), V, sizeof (t)) #define SSF (n) scanf ("%s", N) #define SF (n) scanf (" %d ", &n) #define SFF (A, b) scanf ("%d%d ", &a, &b) #define SFFF (a,b,c) scanf ("%d%d%d ", &a, &b, &c  ) #define PF Printf#define Bug pf ("hi\n") using namespace std; #define INF 0x3f3f3f3f#define n 17ll dp[n];ll K;int len;int A[30];char c[n];int ans[n];void inint () {int i,j;dp[0]=1;dp[1]=1;fre (i,2,n) dp[i]=dp[i-1]*i;} void Dfs (int pos,ll k) {if (Pos==len) return;   printf ("%lld\n", K);   int i,j;   int le=len-pos; Fre (i,0,27) if (a[i]>0) {ll ss=dp[le-1]; Fre (j,0,27) if (A[j]) {if (j==i) {ss/=dp[a[i]-1];} else ss/=dp[a[  J]];  } if (ss>=k)//This bit is I when there is a SS case {ans[pos]=i;  a[i]--;  DFS (POS+1,K);  return; } else K-=ss;}}   int main () {int i,j;   Inint (); while (scanf ("%s%lld", c,&k)) {if (c[0]== ' # ' &&k==0) break; mem (a,0); Len=strlen (c); Fre (I,0,len) a[c[i]-' a '   ]++;  Sort (C,c+len);d FS (0,k), Fre (I,0,len) pf ("%c", ans[i]+ ' A ');  Puts ("");} return 0;}

CSU 1563:lexicography (mathematical counting problem)