HDU 3292 No more tricks, Mr Nanguo 矩阵快速幂求佩尔方程第k大的解

本文深入解析了一个发生在战国时期的故事，讲述了音乐家们如何通过演奏形成正方形排列来取悦国王，并揭示了一个名为《 Bethere just to make up the number》的成语背后的历史背景。通过数学的角度，探讨了正方形排列与音乐之间的联系，以及如何用现代编程技巧解决与之相关的问题。

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No more tricks, Mr Nanguo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 219 Accepted Submission(s): 125

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

Sample Output

   
    7181
600 
No answers can meet such conditions

Author

B.A.C

根据题意知X^2-N*Y^2=1.如果N是完全平方数，则无解，否则先求出特解，在结合快速幂求出第k大的解。

//0MS	228K
#include<stdio.h>
#include<math.h>
#define M 8191
int x,y,n,d;//x,y为最小正整数解
struct Matrax
{
    int m[4][4];
}per,a;
void search(int d)
{
    y=1;
    while(1)
    {
        x=(long long)sqrt(d*y*y+1);
        if(x*x-d*y*y==1)break;
        y++;
    }
}
void init()//建立矩阵
{
    a.m[0][0]=x%M;
    a.m[0][1]=d*y%M;
    a.m[1][0]=y%M;
    a.m[1][1]=x%M;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
            per.m[i][j]=(i==j);
}
Matrax multi(Matrax a,Matrax b)//矩阵相乘
{
    Matrax c;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            c.m[i][j]=0;
            for(int k=0;k<n;k++)
                c.m[i][j]+=a.m[i][k]*b.m[k][j];
            c.m[i][j]%=M;
        }
        return c;
}
Matrax power(int k)//矩阵快速幂
{
    Matrax c,p,ans=per;
    p=a;
    while(k)
    {
        if(k&1){ans=multi(ans,p);k--;}
        else {k/=2;p=multi(p,p);}
    }
    return ans;
}
int main()
{
    int k;
    while(scanf("%d%d",&d,&k)!=EOF)
    {
        int aa=sqrt(d);
        if(aa*aa==d){printf("No answers can meet such conditions\n");continue;}
        search(d);
        n=2;
        init();
        a=power(k-1);
        x=(a.m[0][0]*x%M+a.m[0][1]*y%M)%M;
        printf("%d\n",x);
    }
    return 0;
}