<原># HDU 3292 No more tricks, Mr Nanguo [佩尔方程+矩阵]【数论】

# HDU 3292 No more tricks, Mr Nanguo [佩尔方程+矩阵]【数论】

2016年09月26日 19:21:33 Tabris_ 阅读数：345

https://blog.csdn.net/qq_33184171/article/details/52673118

———————————————————–.
No more tricks, Mr Nanguo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K
(Java/Others)
Total Submission(s): 378 Accepted Submission(s): 250

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

—————————————-.

XX=1+NYY

X
X-NYY=1

X(n)=X(n-1)X(1)+N*Y(n-1)Y(1)
Y(n)=X(n-1)Y(1)+Y(n-1)X(1)

———————————.

``````#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

#define INF 0x3f3f3f3f
#define pb push_back
#define abs(a) (a)>0?(a):-(a)
#define lalal puts("*******");

typedef long long int LL ;
/**********************************/
const int MOD = 8191;
const int M = 2;
struct Matrix
{
LL m[M][M];
void clearO()
{
for(int i=0; i<M; i++) //初始化零矩阵
for(int j=0; j<M; j++)
m[i][j]= 0;
}
void clearE()
{
for(int i=0; i<M; i++) //初始化单位矩阵
for(int j=0; j<M; j++)
m[i][j]= (i==j);
}
void display()
{
for(int i=0; i<M; i++)
{
for(int j=0; j<M; j++)
printf("%d ",m[i][j]);
puts("");
}
}
};

Matrix operator * (Matrix a,Matrix b)
{
Matrix c;
c.clearO();

for(int k=0; k<M; k++)
for(int i=0; i<M; i++) //实现矩阵乘法
{
if(a.m[i][k] == 0)  continue;
for(int j=0; j<M; j++)
{
if(b.m[k][j] ==  0)    continue;
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]+MOD)%MOD;
}
}
return c;
}

Matrix operator ^ (Matrix a,LL b)
{
Matrix c;
c.clearE();
while(b)
{
if(b&1) c= c * a ;
b >>= 1;
a = a * a ;
}
return c;
}
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
if(4==n||9==n||16==n||25==n)
{
puts("No answers can meet such conditions");
continue;
}

LL x,y;

int tmp,tem;
for(int i=1;; i++)
{
tmp=n*i*i+1;
tem=sqrt(tmp);
if(tmp==tem*tem)
{
x=tem,y=i;
break;
}
}

Matrix a,b;
a.clearO(),b.clearO();
a.m[0][0] = x%MOD,a.m[0][1] = y%MOD;

b.m[0][0] = x%MOD,b.m[0][1] = y%MOD;
b.m[1][0] = n*y%MOD,b.m[1][1] = x%MOD;

b=b^(k-1);
a=a*b;

printf("%I64d\n",a.m[0][0]);

//printf("%I64d %I64d\n",x,y);

}
return 0;
}
``````

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