HDU 3292(No more tricks, Mr Nanguo 佩尔方程矩阵快速幂求解)

No more tricks, Mr Nanguo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 298    Accepted Submission(s): 188

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
  

 

Author

B.A.C

 

Source

2010 “HDU-Sailormoon” Programming Contest

 

题目大意:

转化为方程等价于求方程X^2 - N * Y ^ 2 = 1.

解题思路:

若N是完全平方数,则肯定无解,否则用暴力法求出方程的特解,然后根据佩尔方程矩阵快速幂求法得解。

AC代码:

#include<iostream>
#include<cmath>
using namespace std;

int n;
int x,y;
typedef long long ll;
const int maxn = 4;
const int mod = 8191;

typedef struct
{
    int m[maxn][maxn];    
}Matrax;

Matrax a,per;

void sovle() //暴力法求特解 ,x,y为特解 
{
    y = 1;
    while(1)
    {
        x = (ll)sqrt(n * y * y  + 1);
        if(x * x - n * y * y == 1)
        {
            break;
        }
        y++;
    }
}

void init() //矩阵快速幂初始化 
{
    int i,j;
    a.m[0][0] = x % mod;
    a.m[0][1] = n * y % mod;
    a.m[1][0] = y % mod;
    a.m[1][1] = x % mod;
    for(i=0;i<2;i++)
    {
        for(j=0;j<2;j++)
        {
            per.m[i][j] = (i == j);
        }
    }
}

Matrax multi(Matrax a,Matrax b) //矩阵乘法 
{
    Matrax c;
    int k,i,j;
    for(i=0;i<2;i++)
    {
        for(j=0;j<2;j++)
        {
            c.m[i][j] = 0; //初始化 
            for(k=0;k<2;k++)
            {
                c.m[i][j] += a.m[i][k] * b.m[k][j];
            }
            c.m[i][j] %= mod;
        }
    }
    return c;
}

Matrax power(int k) //矩阵乘方 
{
    Matrax p,ans = per;
    p = a;
    while(k)
    {
        if(k & 1)
        {
            ans = multi(ans,p);
            k--;    
        }    
        k /= 2;
        p = multi(p,p);
    }    
    return ans;
}

int main()
{
    int k;
    //freopen("1.txt","r",stdin);
    while(scanf("%d%d",&n,&k) != EOF)
    {
        int yi = sqrt(n + 0.0); 
        if(yi * yi == n) //完全平方数不满足使用佩尔方程 
        {
            printf("No answers can meet such conditions\n");
            continue;
        }
        sovle(); //求一组特解x,y 
        init(); //初始化矩阵快速幂 
        a = power(k-1); //k - 1次方 
        x = (a.m[0][0] * x % mod + a.m[0][1] * y % mod) % mod; //求x 
        printf("%d\n",x);    
    } 
    return 0;
}