UVALive6814 Lexicography

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An anagram of a string is any string that can be formedusing the same letters as the original. (We consider the original string ananagram of itself as well.) For example, the string ACM has thefollowing 6 anagrams, as given in alphabetical order:

ACM

AMC

CAM

CMA

MAC

MCA

As another example, the string ICPC has the following 12anagrams (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 determinethe K^th such anagram according to alphabetical order.

Input: Each test case willbe designated on a single line containing the original word followed by thedesired rank K. Words will use uppercase letters (i.e., A through Z) and willhave length at most 16. The value of K will be in the range from 1 tothe number of distinct anagrams of the given word. A line of the form "# 0"designates the end of the input.

Warning: The value of Kcould be almost 2⁴⁵ in the largest tests, so you should use type long in Java,or type long long in C++ to store K.

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

#include <stdio.h>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#define ll long long
#define N 50

using namespace std;

char str[N];
ll k;
int num[N];
ll f[N];

ll fun(int x)
{
    ll ans=f[x];
    for(int i=0;i<26;i++)
    ans/=f[num[i]];//除以相同个数的阶乘

    return ans;
}

void solve(int len)
{

    for(int i=0;i<len;i++)
    {
        for(int j=0;j<26;j++)
        {
            if(num[j])
            {
                num[j]--;//当前被占用了一个
                ll a=fun(len-i-1);

                num[j]++;
                if(a>=k)
                {
                    num[j]--;
                    printf("%c",'A'+j);
                    break;
                }
                else
                k-=a;
            }
        }
    }
    cout<<endl;

}

int main()
{
    f[0]=1;
    for(int i=1;i<=17;i++)//阶乘
    f[i]=f[i-1]*i;

    while(~scanf("%s %lld",str,&k))
    {
        if(k==0) break;

        int len=strlen(str);
        memset(num,0,sizeof num);

        for(int i=0;i<len;i++)
        {
            num[str[i]-'A']++;
        }
        solve(len);
    }
    return 0;
}

UVALive6814 Lexicography

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原文地址：http://blog.csdn.net/wust_zjx/article/details/45485403