【HDU4919】递推java大数递归优化

最新推荐文章于 2020-12-27 19:27:03 发布

原创最新推荐文章于 2020-12-27 19:27:03 发布 · 786 阅读

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本文详细介绍了如何通过递归和哈希映射技术优化解决特定计算问题，包括输入数据处理、算法实现及性能提升。特别关注了在处理大量数据时，如何有效利用内存减少计算时间，提供了实际应用案例。

Exclusive or

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 391 Accepted Submission(s): 167

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<10 ⁵⁰⁰).

Output

For each tests:

A single integer, the value of the sum.

Sample Input

3
4

Sample Output

6
4

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

现在写java都是先写成c++再对着翻译，先上一份，思路跟标程类似

c++

#define DeBUG
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <sstream>
#include <map>
#include <list>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 0x3f3f3f3f
#define EPS 1e-6
#define TRUE true
#define FALSE false
typedef long long LL;
const double PI = acos(-1.0);
//#pragma comment(linker, "/STACK:102400000,102400000")
inline int sgn(double x)
{
    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);
}
#define N 100005
int dfs(int n)
{
    if (n <= 2)
        return 0;
    if (n == 3)
        return 6;
    int k;
    if (n & 1)
    {
        k = (n - 1) / 2;
        return 4 * dfs(k) + 6 * k;
    }
    else
    {
        k = n / 2;
        return 2 * dfs(k) + 2 * dfs(k - 1) + 4 * k - 4;
    }
}
int contribution(int n,int coe)
{
    int result;
    if(n&1)
    {
        result=(n/2)*6;
    }
    else
    {
        result=n*2-4;
    }
    return result*coe;
}
struct node
{
    int a;
    int b;
};
int main()
{
#ifdef DeBUGs
    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);
#endif
    int n;
    while (scanf("%d", &n) + 1)
    {
        // printf("%d\n",  dfs(n));
        node val;
        val.a=1;
        val.b=0;
        int result=0;
        while(n)
        {
            result=result+contribution(n,val.a)+contribution(n+1,val.b);
            node new_nn;      
            if(n&1)
            {
                n=n/2;
                new_nn.a=val.a*4+val.b*2;
                new_nn.b=val.b*2;
            }
            else
            {
                n=n/2-1;
                new_nn.a=val.a*2;
                new_nn.b=val.a*2+val.b*4;
            }
            val=new_nn;
        }
        printf("%d\n", result);
    }

    return 0;
}

java大数翻译

import java.io.*;
import java.util.Scanner;
import java.math.BigInteger;
class Main
{
    public static class Node
    {
        BigInteger a;
        BigInteger b;
    }
    public static BigInteger get(BigInteger  n, BigInteger coe )
    {
        BigInteger result;
        if ((n.intValue() & 1) == 1)
        {
            result = n.divide(BigInteger.valueOf(2)).multiply(BigInteger.valueOf(6));
        }
        else
        {
            result = n.multiply(BigInteger.valueOf(2)).add(BigInteger.valueOf(-4));
        }
        return result.multiply(coe);
    }
    public static void main(String[]Args) throws FileNotFoundException
    {
        InputStream in=System.in;
        // InputStream in = new FileInputStream(new File("C:\\Users\\Sky\\Desktop\\1.in"));
        Scanner s = new Scanner(in);
        BigInteger n;
        while (s.hasNext())
        {
            n = s.nextBigInteger();
            Node val = new Node();
            val.a = BigInteger.valueOf(1);
            val.b = BigInteger.valueOf(0);
            BigInteger result = BigInteger.ZERO;
            while (n.intValue() != 0)
            {
                result = result.add(get(n, val.a)).add(get(n.add(BigInteger.ONE), val.b));
                Node new_val = new Node();
                if ((n.intValue() & 1) == 1)
                {
                    n = n.divide(BigInteger.valueOf(2));
                    new_val.a = val.a.multiply(BigInteger.valueOf(4)).add(val.b.multiply(BigInteger.valueOf(2)));
                    new_val.b = val.b.multiply(BigInteger.valueOf(2));
                }
                else
                {
                    n = n.divide(BigInteger.valueOf(2)).add(BigInteger.ONE.negate());
                    new_val.a = val.a.multiply(BigInteger.valueOf(2));
                    new_val.b = val.a.multiply(BigInteger.valueOf(2)).add(val.b.multiply(BigInteger.valueOf(4)));
                }
                val=new_val;
            }
            System.out.println(result);
        }
    }
}

直接递归写了发现很慢，后来听锋爷一个hashmap就过了~~O(∩_∩)O~~

c++版

#define DeBUG
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <sstream>
#include <map>
#include <list>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 0x3f3f3f3f
#define EPS 1e-6
#define TRUE true
#define FALSE false
typedef long long LL;
const double PI = acos(-1.0);
//#pragma comment(linker, "/STACK:102400000,102400000")
inline int sgn(double x)
{
    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);
}
#define N 100005
std::map<int, int> mp;
int dfs(int n)
{
    printf("%d\n", n);
    if(mp[n]!=0)
        return mp[n];
    if (n <= 2)
        return 0;
    if (n == 3)
        return 6;
    int k;
    if (n & 1)
    {
        k = (n - 1) / 2;
            mp[n] = 4 * dfs(k) + 6 * k;
        return mp[n];
    }
    else
    {
        k = n / 2;
            mp[n] = 2 * dfs(k) + 2 * dfs(k - 1) + 4 * k - 4;
        return mp[n];
    }
}
int main()
{
#ifdef DeBUGs
    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);
#endif
    int n;
    while (scanf("%d", &n) + 1)
    {
        cout<<dfs(n)<<endl;
    }

    return 0;
}

java版

import java.io.*;
import java.math.BigInteger;
import java.util.HashMap;
import java.util.Scanner;

class Main
{
	public static BigInteger two = BigInteger.valueOf(2);
	public static BigInteger three = BigInteger.valueOf(3);
	public static BigInteger onefu = BigInteger.ONE.negate();
	public static HashMap<BigInteger, BigInteger>hash=new HashMap<BigInteger, BigInteger>();
	public static BigInteger dfs(BigInteger n)
	{
		if(hash.get(n)!=null)
			return hash.get(n);
		if (n.compareTo(two) <= 0)
			return BigInteger.ZERO;
		if (n.compareTo(three) == 0)
			return BigInteger.valueOf(6);
		BigInteger k;
		if ((n.intValue() & 1) == 1)
		{
			k = (n.add(onefu).divide(two));
			BigInteger now=BigInteger.valueOf(4).multiply(dfs(k)).add(BigInteger.valueOf(6).multiply(k));
			hash.put(n, now);
			return now;
		}
		else
		{
			k = n.divide(two);
			BigInteger now=two.multiply(dfs(k)).add(two.multiply(dfs(k.add(onefu))))
					.add(BigInteger.valueOf(4).multiply(k)).add(BigInteger.valueOf(-4));
			hash.put(n, now);
			return now;
		}

	}

	public static void main(String[] Args) throws FileNotFoundException
	{
		InputStream in = System.in;
//		 InputStream in = new FileInputStream(new
//		 File("C:\\Users\\Sky\\Desktop\\1.in"));
		Scanner s = new Scanner(in);
		BigInteger n;
		while (s.hasNext())
		{
			n = s.nextBigInteger();
			System.out.println(dfs(n));
		}
	}
}