csu 1563 Lexicography

1563: Lexicography

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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:

ACM
AMC
CAM
CMA
MAC
MCA
As another example, the string ICPC has the following 12 anagrams (in alphabetical order):

CCIP
CCPI
CICP
CIPC
CPCI
CPIC
ICCP
ICPC
IPCC
PCCI
PCIC
PICC
Given 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 Input

ACM 5ICPC 12REGION 274# 0

Sample Output

MACPICCIGNORE

Hint

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.

Source

这道题就是问一个字符串按照全排列的顺序，输出第几个排列就可以了。首先对字符串按字典序排序，然后按照全排列的规则找规律。

#include<map>#include<set>#include<cmath>#include<stack>#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<iomanip>#include<iostream>#include<algorithm>using namespace std;int main(){char a[111];long long int  m;    long long int jc[22];jc[0]=1;for(int i=1;i<=17;i++)jc[i]=i*jc[i-1];while(cin>>a>>m){if(a[0]=='#'&&m==0)break;sort(a,a+strlen(a));int vis[27];char ans[111];  memset(vis,0,sizeof(vis));int i,j;for(i=0;i<strlen(a);i++)vis[a[i]-'A']++;for(i=0;i<strlen(a);i++){long long  cnt=0;long long  sss=0;for(j=0;j<26;j++){if(vis[j]){   cnt=jc[strlen(a)-1-i];   for(int k=0;k<26;k++)   {   if(k==j)cnt/=jc[vis[k]-1];   else cnt/=jc[vis[k]];   }   if(cnt+sss>=m)   {   ans[i]=j+'A';   vis[j]--;   m-=sss;   break;   }   else sss+=cnt;}}}for(int s=0;s<strlen(a);s++)cout<<ans[s];cout<<endl;}return 0;}