hdu 4919 java大数

Exclusive or

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 277    Accepted Submission(s): 107

Problem Description

Given n, find the value of 


Note: ⊕ denotes bitwise exclusive-or.

 

Input

The input consists of several tests. For each tests:

A single integer n (2≤n<10⁵⁰⁰).

 

Output

For each tests:

A single integer, the value of the sum.

 

Sample Input

34

 

Sample Output

64

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

为java记录一下。

import java.math.BigInteger;import java.io.*;import java.util.*;public class Main {    public static HashMap<BigInteger,BigInteger>  h;    public static BigInteger solve(BigInteger n){        if(h.containsKey(n)) return h.get(n);        if (n.equals(BigInteger.ZERO)) return BigInteger.ZERO;        if (n.equals(BigInteger.ONE)) return BigInteger.ZERO;        BigInteger tp;        BigInteger tp1=n.divide(BigInteger.valueOf(2));        BigInteger er=BigInteger.valueOf(2);                        if(n.mod(BigInteger.valueOf(2)).equals(BigInteger.ONE)){            BigInteger tmp=solve(tp1);            tp=tmp.multiply(BigInteger.valueOf(4)).add(tp1.multiply(BigInteger.valueOf(6)));        }        else {            BigInteger tmp=solve(tp1);            BigInteger tmp1=solve(tp1.subtract(BigInteger.ONE));            tp=tmp.multiply(BigInteger.valueOf(2)).add(tmp1.multiply(BigInteger.valueOf(2)))                    .add(tp1.multiply(BigInteger.valueOf(4))).subtract(BigInteger.valueOf(4));        }        h.put(n,tp);        return tp;    }    public static void main(String[] args){        h=new HashMap<BigInteger, BigInteger>();        Scanner in=new Scanner(System.in);         BigInteger n;        for(int i=2;i<=10000;i++){            solve(BigInteger.valueOf(i));        }        while (in.hasNext()) {            n = in.nextBigInteger();            System.out.println(solve(n));        }    }}