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

2016-09-26 19:21:33 Tabris_ 阅读数：375

博客爬取于2020-06-14 22:43:09
以下为正文

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

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=3292

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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): 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

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题目大意：
南郭先生的故事想必大家都知道
就是最开始有X*X人
后来南郭走了
变成了N个Y*Y
之后题目给你一个N 和K
让你求满足题意的第K大的X值是多少。。。

解题思路：
题目显然能够得到下面的方程
XX=1+NYY
转化一下就是
XX-NYY=1

这就是标准的佩尔方程了 不知道佩尔方程的话可以先去百度下.
首先N很小 所以可以先暴力的跑出特解 然后用迭代的方式求得第K大的解
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)

因为K很大 所以要用矩阵优化一下

矩阵略

附本题代码
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# 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;
}