[JAVA]HDU 4919 Exclusive or

题意很简单, 就是给个n, 算下面这个式子的值.

$\sum\limits_{i=1}^{n-1} i \otimes (n-i)$

重点是n的范围:2≤n<10⁵⁰⁰

比赛的时候 OEIS一下得到了一个公式:

$a_0=a_1=a_2=0$;

n为偶数 : $2 \times a_{\frac{n}{2}}+2 \times a_{\frac{n}{2}-1}+4\times (\frac{n}{2}-1) $

n为奇数 : $4\times a_{\frac{n-1}{2}}+6\times\frac{n-1}{2}$

然后勇敢的打了一发暴力...想也知道肯定TLE...

之后 学到了一种机智的按位算的方法

 import java.io.*;
 import java.util.*;
 import java.math.*;

 public class Main
 {
     static BigInteger yi=BigInteger.ONE;
     static BigInteger er=BigInteger.valueOf(2);
     static BigInteger li=BigInteger.ZERO;
     public static void main(String[] args)
     {
         InputReader in = new InputReader();
         PrintWriter out = new PrintWriter(System.out);
         BigInteger []bit=new BigInteger[2005];
         bit[0]=yi;
         for(int i=1; i<=2000; i++)
             bit[i]=bit[i-1].multiply(er);
         while(in.hasNext())
         {
             BigInteger n=new BigInteger(in.next());
             int []wei=new int[2005];
             int d=0;
             BigInteger []a=new BigInteger[2005];
             BigInteger []b=new BigInteger[2005];
             BigInteger tmp=n;
             while(tmp.compareTo(li)!=0)
             {
                 if(tmp.mod(er).equals(li))
                     wei[d++]=0;
                 else
                     wei[d++]=1;
                 tmp=tmp.divide(er);
             }
             BigInteger sum=li, ji=yi;
             for(int i=0; i<d; i++)
             {
                 if(wei[i]>0)
                     sum=sum.add(ji);
                 ji=ji.multiply(er);
                 a[i+1]=sum;
             }
             sum=li;
             for(int i=d; i>=0; i--)
             {
                 sum=sum.multiply(er).add(BigInteger.valueOf(wei[i]));
                 b[i+1]=sum;
             }
             a[0]=li;
             BigInteger ans=li;
             for(int i=0; i<d; i++)
                 if(wei[i]==0)
                 {
                     BigInteger an=(bit[i].subtract(a[i]).subtract(yi)).multiply(b[i+2]).multiply(bit[i]);
                     ans=ans.add(an).add(an);
                 }
                 else
                 {
                     BigInteger an=((a[i].add(yi)).multiply(b[i+2].add(yi)).subtract(yi)).multiply(bit[i]);
                     ans=ans.add(an).add(an);
                 }
             out.println(ans);
         }
         out.close();
     }
 }
 class InputReader
 {
     BufferedReader buf;
     StringTokenizer tok;
     InputReader()
     {
         buf = new BufferedReader(new InputStreamReader(System.in));
     }
     boolean hasNext()
     {
         while(tok == null || !tok.hasMoreElements())
         {
             try
             {
                 tok = new StringTokenizer(buf.readLine());
             }
             catch(Exception e)
             {
                 return false;
             }
         }
         return true;
     }
     String next()
     {
         if(hasNext())
             return tok.nextToken();
         return null;
     }
     int nextInt()
     {
         return Integer.parseInt(next());
     }
     long nextLong()
     {
         return Long.parseLong(next());
     }
     double nextDouble()
     {
         return Double.parseDouble(next());
     }
     BigInteger nextBigInteger()
     {
         return new BigInteger(next());
     }
     BigDecimal nextBigDecimal()
     {
         return new BigDecimal(next());
     }
 }

HDOJ 4919

再之后 学习了一下map的记忆化搜索

 import java.io.*;
 import java.util.*;
 import java.math.*;

 public class Main
 {
     static BigInteger yi=BigInteger.ONE;
     static BigInteger er=BigInteger.valueOf(2);
     static BigInteger li=BigInteger.ZERO;
     static BigInteger sa=BigInteger.valueOf(3);
     static BigInteger si=BigInteger.valueOf(4);
     static BigInteger liu=BigInteger.valueOf(6);
     static HashMap<BigInteger, BigInteger> a=new HashMap<BigInteger, BigInteger>();
     public static BigInteger dfs(BigInteger n)
     {
         if(a.containsKey(n))
             return a.get(n);
         BigInteger m;
         if(n.mod(er).equals(li))
         {
             BigInteger aa=n.divide(er);
             BigInteger bb=dfs(aa).multiply(er);
             BigInteger cc=dfs(aa.subtract(yi)).multiply(er);
             m=(aa.subtract(yi)).multiply(si).add(bb).add(cc);
         }
         else
         {
             BigInteger aa=(n.subtract(yi)).divide(er);
             m=dfs(aa).multiply(si).add(aa.multiply(liu));
         }
         a.put(n, m);
         return m;
     }
     public static void main(String[] args)
     {
         InputReader in = new InputReader();
         PrintWriter out = new PrintWriter(System.out);
         a.put(li, li);
         a.put(yi, li);
         a.put(er, li);
         while(in.hasNext())
         {
             BigInteger n=new BigInteger(in.next());
             out.println(dfs(n));
         }
         out.close();
     }
 }
 class InputReader
 {
     BufferedReader buf;
     StringTokenizer tok;
     InputReader()
     {
         buf = new BufferedReader(new InputStreamReader(System.in));
     }
     boolean hasNext()
     {
         while(tok == null || !tok.hasMoreElements())
         {
             try
             {
                 tok = new StringTokenizer(buf.readLine());
             }
             catch(Exception e)
             {
                 return false;
             }
         }
         return true;
     }
     String next()
     {
         if(hasNext())
             return tok.nextToken();
         return null;
     }
     int nextInt()
     {
         return Integer.parseInt(next());
     }
     long nextLong()
     {
         return Long.parseLong(next());
     }
     double nextDouble()
     {
         return Double.parseDouble(next());
     }
     BigInteger nextBigInteger()
     {
         return new BigInteger(next());
     }
     BigDecimal nextBigDecimal()
     {
         return new BigDecimal(next());
     }
 }

HDOJ 4919