1563: LexicographyTime Limit: 1 SecMemory Limit: 128 MBSubmit: 342Solved: 111[Submit][Status][Web Board]Description
An anagram of a string is 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 has the following 6 anagrams, as given in alphabetical order:
ACMAMCCAMCMAMACMCAAs another example, the string ICPC has the following 12 anagrams (in alphabetical order):
CCIPCCPICICPCIPCCPCICPICICCPICPCIPCCPCCIPCICPICCGiven a string and a rank K, you are to determine the Kth such anagram according to alphabetical order.
Input
Each test case will be designated on a single line containing the original word followed by the desired rank K. Words will use uppercase letters (i.e., A through Z) and will have length at most 16. The value of K will 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 InputACM 5ICPC 12REGION 274# 0Sample OutputMACPICCIGNOREHINT
The value of K could be almost 245 in the largest tests, so you should use type long in Java, or type long long in C++ to store K.
//排列组合计数问题，n个数全排列为n!,如果x有t个，那么全排列为n!/(t!)//然后枚举当前位是i的情况，，然后一直减去为i的时候的情况，如果<i的情况，//那么这一位就是i了#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 pfprintf#define bugpf("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) //这一位是i时有ss种情况{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);dfs(0,k);fre(i,0,len)pf("%c",ans[i]+'A');puts(""); } return 0;}
君子无故，玉不去身。
