hdu 3292 No more tricks, Mr Nanguo

No more tricks, Mr Nanguo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 494    Accepted Submission(s): 334

Problem Description

Now
Sailormoon girls want to tell you a ancient idiom story named “be there
just to make up the number”. The story can be described by the
following words.
In the period of the Warring States (475-221 BC),
there was a state called Qi. The king of Qi was so fond of the yu, a
wind instrument, that he had a band of many musicians play for him every
afternoon. The number of musicians is just a square number.Beacuse a
square formation is very good-looking.Each row and each column have X
musicians.
The king was most satisfied with the band and the
harmonies they performed. Little did the king know that a member of the
band, Nan Guo, was not even a musician. In fact, Nan Guo knew nothing
about the yu. But he somehow managed to pass himself off as a yu player
by sitting right at the back, pretending to play the instrument. The
king was none the wiser. But Nan Guo's charade came to an end when the
king's son succeeded him. The new king, unlike his father, he decided to
divide the musicians of band into some equal small parts. He also wants
the number of each part is square number. Of course, Nan Guo soon
realized his foolish would expose, and he found himself without a band
to hide in anymore.So he run away soon.
After he leave,the number of
band is Satisfactory. Because the number of band now would be divided
into some equal parts,and the number of each part is also a square
number.Each row and each column all have Y musicians.

 

Input

There
are multiple test cases. Each case contains a positive integer N ( 2
<= N < 29). It means the band was divided into N equal parts. The
folloing number is also a positive integer K ( K < 10^9).

 

Output

There
may have many positive integers X,Y can meet such conditions.But you
should calculate the Kth smaller answer of X. The Kth smaller answer
means there are K – 1 answers are smaller than them. Beacuse the answer
may be very large.So print the value of X % 8191.If there is no answers
can meet such conditions,print “No answers can meet such conditions”.

 

Sample Input

2 999888

3 1000001

4 8373

 

Sample Output

7181

600

No answers can meet such conditions

依题可得x^2-ny^2=1,所以此题解法为佩尔方程+矩阵快速幂

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#pragma comment(linker, "/stck:1024000000,1024000000")
#define lowbit(x) (x&(-x))
#define max(x,y) (x>=y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.1415926535897932384626433832
#define ios() ios::sync_with_stdio(true)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll n,k,x,y;
const ll maxn=;
struct matrix
{
    ll a[][];
};
void serach(ll n,ll &x,ll &y)
{
    y=;
    while()
    {
        x=(1ll)*sqrt(n*y*y+);
        if(x*x-n*y*y==) break;
        y++;
    }
}
matrix mulitply(matrix ans,matrix pos)
{
    matrix res;
    memset(res.a,,sizeof(res.a));
    for(int i=;i<;i++)
    {
        for(int j=;j<;j++)
        {
            for(int k=;k<;k++)
            {
                res.a[i][j]+=(ans.a[i][k]*pos.a[k][j])%maxn;
                res.a[i][j]%=maxn;
            }
        }
    }
    return res;
}
matrix quick_pow(ll m)
{
    matrix ans,pos;
    for(int i=;i<;i++)
        for(int j=;j<;j++)
            ans.a[i][j]=(i==j);
    pos.a[][]=x%maxn;
    pos.a[][]=n*y%maxn;
    pos.a[][]=y%maxn;
    pos.a[][]=x%maxn;
    while(m)
    {
        if(m&) ans=mulitply(ans,pos);
        pos=mulitply(pos,pos);
        m>>=;
    }
    return ans;
}
int main()
{
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        ll m=sqrt(n);
        if(m*m==n) {printf("No answers can meet such conditions\n");continue;}
        serach(n,x,y);
        matrix ans=quick_pow(k);
        printf("%lld\n",ans.a[][]);
    }
    return ;
}