SDUT 3185 Lexicography（排列问题（不会））

Lexicography

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

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 K^th such anagram according
to alphabetical order.

输入

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.

Warning: 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.

输出

For each test, display the K^th anagram
of the original string.

示例输入

ACM 5
ICPC 12
REGION 274
# 0

示例输出

MAC
PICC
IGNORE

提示

来源

2014 ACM MId-Central Reginal Programming Contest（MCPC2014）

示例程序

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
#define LL long long
using namespace std;
LL a[20];
void init()
{   int i;
    a[0]=1;
    for(i=1;i<=17;i++)
    {
        a[i]=a[i-1]*i;
    }
}
string se(string str,LL  n)
{   int i;
    string s;
    map<int,int>st;
    int l=str.length();
    for(i=0;i<=l-1;i++)
    {
        st[str[i]-'A']++;
    }

    while(st.size())
    {
        l--;
        LL sum=0;
    for(map<int,int>::iterator it=st.begin();it!=st.end();it++)
    {
        LL ans=a[l]/a[it->second-1];
        for(map<int,int>::iterator j=st.begin();j!=st.end();j++)
        {
            if(it!=j)
            {
                ans=ans/a[j->second];
            }
        }
        if(sum+ans>=n)
        {
          s+=char(it->first+'A');
          it->second--;
          if(it->second==0)
          {
              st.erase(it);
          }
            n=n-sum;
            break;
        }
        sum+=ans;
    }
    }
    return s;
}
int main()
{
    init();
    string str;
      LL n;
    while(cin>>str>>n)
    {
        if(str=="#"&&n==0)
        {
            break;
        }
        cout<<se(str,n)<<endl;
    }
    return 0;
}