Palindromic Permutation - UVa 11027 组合数

最新推荐文章于 2025-11-04 15:05:03 发布

原创最新推荐文章于 2025-11-04 15:05:03 发布 · 411 阅读

1 ·

CC 4.0 BY-SA版权

dp动态规划和递推同时被 2 个专栏收录

281 篇文章

订阅专栏

UVa

218 篇文章

订阅专栏

本文介绍了一种算法，用于给定一个字符串后构造出字典序第n个回文串的方法。通过分析字符串中各字符出现次数，利用排列组合原理来确定回文串的可行性及构造过程。

Given a string of characters, we can permute the individual characters to make new strings. We can then order these strings into alphabetical order.

For example the string ‘abba’ gives rise to the following 6 distinct permutations in alphabetical order.

aabb 1

abab 2

abba 3

baab 4

baba 5

bbaa 6

Of these 6 permutations, only 2 are palindromes (A string that reads the same when read backwards). These are ‘abba’ and ‘baab’.

Given a string, you have to find out the nth palindrome in the sorted list of all permutations. For the above case ‘abba’ is the 1^st and ‘baab’ is the 2^nd palindrome.

Input

The first line of input gives the number of test cases. Each case contains a string, consisting of lowercase letters only, followed by a space separated positive integer n (n < 2^31). The length of the string will be at most 30.

Output

For each case, output the case number followed by the nth palindrome, but if the total number of palindromes is less than n output ‘XXX’ without the quotes. Follow the sample for exact format.

Sample Input

Output for Sample Input

abba 1

abba 2

abba 3

Case 1: abba

Case 2: baab

Case 3: XXX

题意：给你一个字符串，让你构造字典序第n的回文串。

思路：假设一个偶数的回文串，那么它左半边的字母的组合情况，就是这个字符串可以变成的回文串的个数。先判断是否可以形成回文串，然后从左到右依次构造左半边的回文串序列，假设去掉a后剩下的可以构造k1个字符串，而k1>=m那么这个位置就可以放a，否则m-=k1，再去尝试b。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,n,t,num[30];
ll m,c[30][30];
char s[40],ans[40];
ll solve(int len)
{
    int i;
    ll temp=1;
    for(i=0;i<26;i++)
    {
        temp*=c[len][num[i]];
        len-=num[i];
    }
    return temp;
}
bool check()
{
    int i,k=0;
    for(i=0;i<26;i++)
    {
        if(num[i]&1)
        {
           k++;
           ans[n/2+1]='a'+i;
        }
        num[i]/=2;
    }
    if(k>1)
      return false;
    return solve(n/2)>=m;
}
void dfs(int pos)
{
    if(pos>n/2)
      return;
    int i,j,k;
    ll ret;
    for(i=0;i<26;i++)
       if(num[i]>0)
       {
           num[i]--;
           ret=solve(n/2-pos);
           if(ret>=m)
           {

               ans[pos]='a'+i;
               dfs(pos+1);
               return ;
           }
           m-=ret;
           num[i]++;
       }
}
int main()
{
    int i,j,k;
    for(i=0;i<=20;i++)
       c[0][i]=1;
    for(i=1;i<=20;i++)
    {
        c[i][0]=c[i][i]=1;
        for(j=1;j<i;j++)
           c[i][j]=c[i-1][j-1]+c[i-1][j];
    }
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%s%lld",s+1,&m);
        memset(num,0,sizeof(num));
        n=strlen(s+1);
        for(i=1;i<=n;i++)
          num[s[i]-'a']++;
        if(m==0 || !check())
        {
            printf("Case %d: XXX\n",t);
            continue;
        }
        dfs(1);
        printf("Case %d: ",t);
        for(i=1;i<=n/2;i++)
           printf("%c",ans[i]);
        if(n&1)
           printf("%c",ans[n/2+1]);
        for(i=n/2;i>=1;i--)
           printf("%c",ans[i]);
        printf("\n");
    }
}