P2504洗牌(扩欧)

洗牌

根据题目模拟了几次洗牌过程，得到了第一层逻辑，$if(x<=(n/2))x1=2\ast xelseif(x>(n/2))x1=(x-(n/2)-1)\ast 2+1;->x1=(2\ast x-n-2)+1->x1=(2\ast x-(n+1))因为2\ast x-（n+1）<n所以x1=2\ast x(modn+1),总式子也是x1=2\ast x(modn+1)题目变为2^m\ast x=L(modn+1)->2^m\ast x+(n+1)\ast Y=L$ 扩欧取最小值即可

//package KMP;
import java.util.Scanner;
import java.util.Map;
import java.util.Set;
import java.util.TreeMap;
import java.util.LinkedList;
import java.util.Queue;
import java.math.BigInteger;
public class Main {
static long x,y;
public static long exd_gcd(long a,long b)
{
	if(b==0)
	{
		x=1;y=0;
		return a;
	}
	long d=exd_gcd(b,a%b);
	y^=x;
	x^=y;
	y^=x;
	y-=(a/b)*x;
	return d;
}
public static long res(long s,long mod)
{
	exd_gcd(s,mod);
	//System.out.println("X"+x+"Y"+y);
	return x>0?x%mod:(x%mod+mod)%mod;
}
public static long mul(long a,long b,long mod)
{
	long ans=0;
	while(b!=0)
	{
		if((b&1)==1)
			ans=(ans+a)%mod;
		a=(a+a)%mod;
		b>>=1;
	}
	return ans;
}
public static long quicks(long a,long b,long mod)
{
	long ans=1;
	while(b!=0)
	{
		if((b&1)==1)
			ans=mul(ans,a,mod);
		a=mul(a,a,mod);
		b>>=1;
	}
	//System.out.println("A::"+ans);
	return ans;
}
//N M L 长度 洗牌次数 访问的位置
//2^M*X=L(mod N+1)  2^M*x+(N+1)*Y=L
public static void main(String []args)
{
	Scanner cin=new Scanner (System.in);	
	long N,M,L;
	N=cin.nextLong();
	M=cin.nextLong();
	L=cin.nextLong();
	long A=quicks(2,M,N+1);
	long B=N+1;
	//System.out.println(A);
	long X=res(A,B);
	X=mul(X,L,B)%B;
	System.out.println(X);
	cin.close();
}
}