semipal.in / semipal.out
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.
Input
On the first line of the file semipal.in, there is an integer 
(
) representing the number of test cases. On the next lines there are 2 numbers, 
(
and K 
where 
is the number of semipalindromes of length .
Output
In the output file semipal.out, there should be lines, the 
-th of which should contain the answer for the -th test.
Example
2
5 1
5 14
aaaaa
bbabb
显然只需要保证开头和结尾字母相同，就一定是合法的啦。
然后就把中间的部分得到即可。
#include<cstdio>
using namespace std;
typedef long long ll;
int T,n;
ll m;
char a[70];
int main()
{
// freopen("l.in","r",stdin);
freopen("semipal.in","r",stdin);
freopen("semipal.out","w",stdout);
scanf("%d",&T);
for(;T;--T)
{
scanf("%d%I64d",&n,&m);
// ll all=1ll<<(n-1);
bool flag=0;
if(m>(1ll<<(n-2)))
{
m-=(1ll<<(n-2));
flag=1;
}
--m;
for(int i=1;i<=n-2;++i)
{
a[i]=(m%2 ? 'b' : 'a');
m/=2;
}
if(flag)
{
putchar('b');
for(int i=n-2;i>=1;--i)
putchar(a[i]);
puts("b");
}
else
{
putchar('a');
for(int i=n-2;i>=1;--i)
putchar(a[i]);
puts("a");
}
}
return 0;
}