No more tricks, Mr Nanguo HDU - 3292
题意：输入D,K ，求 的第K大X解
用佩尔方程的矩阵表示，马上会总结到（佩尔方程专题总结）
用JAVA写失败了。不知道问题出在哪里
#include<cstdio>
#include<vector>
#include<cmath>
#include<math.h>
#include<time.h>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1005;
const int MOD=8191;
const double eps=1e-10;
typedef struct{
ll mat[20][20];
}matrix;
// ¾ØÕó³Ë·¨
matrix matrix_mul(matrix a,matrix b,ll y){
matrix c;
memset(c.mat,0,sizeof(c.mat));
for(ll i=0;i<=y;i++){
for(ll j=0;j<=y;j++){
for(ll k=0;k<=y;k++){
c.mat[i][j]+=(a.mat[i][k]%MOD*b.mat[k][j]%MOD)%MOD;
c.mat[i][j]%=MOD;
}
}
}
return c;
}
// ¿ìËÙÃÝ
matrix matrix_quick_power(matrix a,ll k ,ll y){
matrix b;
memset(b.mat,0,sizeof(b.mat));
for(int i=0;i<=y;i++)
b.mat[i][i]=1ll;
while(k){
if(k%2==1) b=matrix_mul(b,a,y);
k=k/2;
a=matrix_mul(a,a,y);
}
return b;
}
ll d,k;
void work(ll &x1,ll &y1){
y1=1;
while(true){
ll x_2=1+d*y1*y1;
if(x_2>0){
x1=sqrt(x_2);
if(x1*x1-d*y1*y1==1) return ;
}
++y1;
}
}
bool is_square(ll x){
ll t=sqrt(x);
for(ll i=t-1;i<=t+1;i++)
if(i*i==x) return true;
return false;
}
int main(void){
while(scanf("%lld%lld",&d,&k)==2){
matrix a,res;
ll x1,y1;
if(is_square(d)){
printf("No answers can meet such conditions\n");
continue;
}
work(x1,y1);
a.mat[1][1]=x1;
a.mat[2][1]=y1;
a.mat[1][2]=d*y1;
a.mat[2][2]=x1;
res=matrix_quick_power(a,k-1,5);
ll ansx;
ansx=res.mat[1][1]*x1%MOD+res.mat[1][2]*y1%MOD;
ansx%=MOD;
printf("%lld\n",ansx);
}
return 0;
}