链接:

https://vjudge.net/problem/Gym-100923L

题意:

Por Costel the pig, our programmer in-training, has recently returned from the Petrozaporksk training camp. There, he learned a lot of things: how to boil a cob, how to scratch his belly using his keyboard, etc... He almost remembers a programming problem too:

A semipalindrome is a word for which there exists a subword such that is a prefix of and (reverse ) is a suffix of . For example, 'ababba' is a semipalindrom because the subword 'ab' is prefix of 'ababba' and 'ba' is suffix of 'ababba'.

Let's consider only semipalindromes that contain letters 'a' and 'b'. You have to find the -th lexicographical semipalindrome of length .

Por Costel doesn't remember if the statement was exactly like this at Petrozaporksk, but he finds this problem interesting enough and needs your help to solve it.

思路:

考虑首字母等于末字母,中间按字典序处理即可,字典序处理可用二进制的思想.
aaa = 000, aab = 001, aba == 010, 就是挨个进位,注意,第k小在字典序的二进制中是k-1, 因为aaa = 000,从0开始.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

map<string, double> Sc;
const int MAXN = 1e3+10;
char s[100];
LL n, k;
//aaaa, aaba, abaa, abba, baba,
int main()
{
    freopen("semipal.in", "r", stdin);
    freopen("semipal.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        cin >> n >> k;
        if (k > 1LL<<(n-2))
        {
            k -= 1LL<<(n-2);
            k--;
            s[0] = 'b';
            s[n-1] = 'b';
            for (int i = n-2;i >= 1;i--)
            {
                if (k%2 == 0)
                    s[i] = 'a';
                else
                    s[i] = 'b';
                k /= 2;
            }
        }
        else
        {
            s[0] = 'a';
            s[n-1] = 'a';
            k--;
            for (int i = n-2;i >= 1;i--)
            {
                if (k%2 == 0)
                    s[i] = 'a';
                else
                    s[i] = 'b';
                k /= 2;
            }
        }
        s[n] = 0;
        cout << s << endl;
    }

    return 0;
}