No more tricks, Mr Nanguo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 320 Accepted Submission(s): 207

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

//
//  main.cpp
//  hdu3292
//
//  Created by 陈加寿 on 15/12/1.
//  Copyright (c) 2015年 陈加寿. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <string>
#include <math.h>
using namespace std;

#define MOD 8191

void matrix_mul(long long s[4][4],long long t[4][4])
{
    long long tmp[4][4];
    memset(tmp,0,sizeof(tmp));
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            for(int k=0;k<2;k++)
                tmp[i][j]=(tmp[i][j]+s[i][k]*t[k][j])%MOD;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
            s[i][j]=tmp[i][j];
}

int main(int argc, const char * argv[]) {
    long long n,k;
    while(cin>>n>>k)
    {
        long long x=sqrt( (double)n );
        if(x*x==n)
        {
            printf("No answers can meet such conditions
");
            continue;
        }
        //然后开始寻找第一个解
        long long x0,y0;
        for(long long i=1;;i++)
        {
            x=sqrt((double)(i*i*n+1));
            if(x*x==i*i*n+1)
            {
                x0=x;
                y0=i;
                break;
            }
        }
        long long mat[4][4],ans[4][4];
        mat[0][0]=x0; mat[0][1]=y0;
        mat[1][0]=n*y0; mat[1][1]=x0;
        memset(ans,0,sizeof(ans));
        ans[0][0]=1; ans[1][1]=1;
        k-=1;
        while(k)
        {
            if(k&1) matrix_mul(ans,mat);
            k>>=1;
            matrix_mul(mat,mat);
        }
        long long ans1;
        ans1 = (x0*ans[0][0]+y0*ans[1][0])%MOD;
        cout<<ans1<<endl;
    }
    return 0;
}