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
因为卡内存，所以不能把答案的表全打出来，但是可以每隔100记录一次答案，这样只需要开10w的数组。然后每次询问的时候，从最近的记录的答案开始暴力，不超过100次就能得到答案。
#include<cstdio>
using namespace std;
#define MOD 10000003
typedef long long ll;
int n,a,b,x1,q,q1;
int anss[100010];
int main()
{
freopen("pocnitoare.in","r",stdin);
freopen("pocnitoare.out","w",stdout);
// freopen("k.in","r",stdin);
scanf("%d%d%d%d%d%d",&n,&a,&b,&x1,&q,&q1);
int now=x1;
anss[1]=now;
for(int i=2;i<=10000003;++i)
{
now=(int)((((ll)now*(ll)(i-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
if(i%100==1)
anss[i/100+1]=now;
}
// int now=anss[q1/100+1];
// int tmp=q1%100-1;
// for(int i=1;i<=tmp;++i)
// now=(int)((((ll)now*(ll)(i-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
// printf("%d\n",now);
for(int i=1;i<=q;++i)
{
if(i!=1)
q1=((int)((ll)(i-1)*(ll)now%(ll)MOD)+b%MOD)%MOD+1;
now=anss[(q1-1)/100+1];
for(int j=(q1-1)/100*100+2;j<=q1;++j)
now=(int)((((ll)now*(ll)(j-1)%(ll)n)%(ll)n+(ll)a%(ll)n)%(ll)n);
printf("%d\n",now);
}
return 0;
}