POJ 2506 Tiling

这题是一个标准递推，挺简单的，就是因为有大数操作写个博客记录一下。

递推公式是a[n]=a[n-1]+2*a[n-2]。

下面是代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <limits>
#include <cstdlib>

using namespace std;

const int MAXD = 1000, DIG = 9, BASE = 1000000000;

const unsigned long long BOUND = numeric_limits <unsigned long long> :: max () - (unsigned long long) BASE * BASE;

class bignum
{
private:
    int digits[MAXD];
    int D;
public:

    friend ostream &operator<<(ostream &out,bignum &c);

    inline void trim()
    {
        while(D > 1 && digits[D-1] == 0 )
            D--;
    }

    inline void dealint(long long x)
    {
        memset(digits,0,sizeof(digits));
        D = 0;
        do
        {
            digits[D++] = x % BASE;
            x /= BASE;
        }
        while(x > 0);
    }

    inline void dealstr(char *s)
    {
        memset(digits,0,sizeof(digits));
        int len = strlen(s),first = (len + DIG -1)%DIG + 1;

        D = (len+DIG-1)/DIG;

        for(int i = 0; i < first; i++)
            digits[D-1] = digits[D-1]*10 + s[i] - '0';

        for(int i = first, d = D-2; i < len; i+=DIG,d--)
            for(int j = i; j < i+DIG; j++)
                digits[d] = digits[d]*10 + s[j]-'0';

        trim();
    }

    inline char *print()
    {
        trim();

        char *cdigits = new char[DIG * D + 1];

        int pos = 0,d = digits[D-1];

        do
        {
            cdigits[pos++] = d % 10 + '0';
            d/=10;
        }
        while(d > 0);

        reverse(cdigits,cdigits+pos);

        for(int i = D - 2; i >= 0; i--,pos += DIG)
            for(int j = DIG-1,t = digits[i]; j >= 0; j--)
            {
                cdigits[pos+j] = t%10 + '0';
                t /= 10;
            }

        cdigits[pos] = '\0';

        return cdigits;
    }


    bignum()
    {
        dealint(0);
    }

    bignum(long long x)
    {
        dealint(x);
    }

    bignum(int x)
    {
        dealint(x);
    }

    bignum(char *s)
    {
        dealstr(s);
    }



    inline bool operator < (const bignum &o) const
    {
        if(D != o.D)
            return D < o.D;

        for(int i = D-1; i>=0; i--)
            if(digits[i] != o.digits[i])
                return digits[i] < o.digits[i];
        return false; //equal

    }

    bool operator >  (const bignum & o)const
    {
        return o < *this;
    }
    bool operator <= (const bignum & o)const
    {
        return !(o < *this);
    }
    bool operator >= (const bignum & o)const
    {
        return !(*this < o);
    }
    bool operator != (const bignum & o)const
    {
        return o < *this || *this < o;
    }
    bool operator == (const bignum & o)const
    {
        return !(o < *this) && !(*this < o);
    }


    bignum &operator++()
    {
        *this = *this  + 1;
        return *this;
    }


    bignum operator ++(int)
    {
        bignum old = *this;
        ++(*this);
        return old;
    }

    inline bignum operator << (int p) const
    {
        bignum temp;
        temp.D = D + p;

        for (int i = 0; i < D; i++)
            temp.digits [i + p] = digits [i];

        for (int i = 0; i < p; i++)
            temp.digits [i] = 0;

        return temp;
    }

    inline bignum operator >> (int p) const
    {
        bignum temp;
        temp.D = D - p;

        for (int i = 0; i < D - p; i++)
            temp.digits [i] = digits [i + p];

        for (int i = D - p; i < D; i++)
            temp.digits [i] = 0;

        return temp;
    }


    bignum &operator += (const bignum &b)
    {
        *this = *this + b;
        return *this;
    }

    bignum &operator -= (const bignum &b)
    {
        *this = *this - b;
        return *this;
    }

    bignum &operator *= (const bignum &b)
    {
        *this = *this * b;
        return *this;
    }

    bignum &operator /= (const bignum &b)
    {
        *this = *this / b;
        return *this;
    }

    bignum &operator %= (const bignum &b)
    {
        *this = *this % b;
        return *this;
    }

    inline bignum operator + (const bignum &o) const
    {
        bignum sum = o;
        int carry = 0;

        for (sum.D = 0; sum.D < D || carry > 0; sum.D++)
        {
            sum.digits [sum.D] += (sum.D < D ? digits [sum.D] : 0) + carry;

            if (sum.digits [sum.D] >= BASE)
            {
                sum.digits [sum.D] -= BASE;
                carry = 1;
            }
            else
                carry = 0;
        }

        sum.D = max (sum.D, o.D);
        sum.trim ();
        return sum;
    }
    inline bignum operator - (const bignum &o) const
    {
        bignum diff = *this;

        for (int i = 0, carry = 0; i < o.D || carry > 0; i++)
        {
            diff.digits [i] -= (i < o.D ? o.digits [i] : 0) + carry;

            if (diff.digits [i] < 0)
            {
                diff.digits [i] += BASE;
                carry = 1;
            }
            else
                carry = 0;
        }

        diff.trim ();
        return diff;
    }
    inline bignum operator * (const bignum &o) const
    {
        bignum prod = 0;
        unsigned long long sum = 0, carry = 0;

        for (prod.D = 0; prod.D < D + o.D - 1 || carry > 0; prod.D++)
        {
            sum = carry % BASE;
            carry /= BASE;

            for (int j = max (prod.D - o.D + 1, 0); j <= min (D - 1, prod.D); j++)
            {
                sum += (unsigned long long) digits [j] * o.digits [prod.D - j];

                if (sum >= BOUND)
                {
                    carry += sum / BASE;
                    sum %= BASE;
                }
            }

            carry += sum / BASE;
            prod.digits [prod.D] = sum % BASE;
        }

        prod.trim ();
        return prod;
    }
    inline bignum range (int a, int b) const
    {
        bignum temp = 0;
        temp.D = b - a;

        for (int i = 0; i < temp.D; i++)
            temp.digits [i] = digits [i + a];

        return temp;
    }


    inline double double_div (const bignum &o) const
    {
        double val = 0, oval = 0;
        int num = 0, onum = 0;

        for (int i = D - 1; i >= max (D - 3, 0); i--, num++)
            val = val * BASE + digits [i];

        for (int i = o.D - 1; i >= max (o.D - 3, 0); i--, onum++)
            oval = oval * BASE + o.digits [i];

        return val / oval * (D - num > o.D - onum ? BASE : 1);
    }

    inline pair <bignum, bignum> divmod (const bignum &o) const
    {
        bignum quot = 0, rem = *this, temp;

        for (int i = D - o.D; i >= 0; i--)
        {
            temp = rem.range (i, rem.D);
            int div = (int) temp.double_div (o);
            bignum mult = o * div;

            while (div > 0 && temp < mult)
            {
                mult = mult - o;
                div--;
            }

            while (div + 1 < BASE && !(temp < mult + o))
            {
                mult = mult + o;
                div++;
            }

            rem = rem - (o * div << i);

            if (div > 0)
            {
                quot.digits [i] = div;
                quot.D = max (quot.D, i + 1);
            }
        }

        quot.trim ();
        rem.trim ();
        return make_pair (quot, rem);
    }

    inline bignum operator / (const bignum &o) const
    {
        return divmod (o).first;
    }

    inline bignum operator % (const bignum &o) const
    {
        return divmod (o).second;
    }


    inline bignum power (int exp) const
    {
        bignum p = 1, temp = *this;

        while (exp > 0)
        {
            if (exp & 1) p = p * temp;
            if (exp > 1) temp = temp * temp;
            exp >>= 1;
        }

        return p;
    }

    inline bignum factorial() const
    {
        bignum ans = 1, num = *this;

        if (num == 0 || num == 1)
            return ans;
        while (!(num < 0 || num == 0))
        {
            ans = ans * num;
            num = num - 1;
        }
        return ans;
    }
};

ostream &operator<<(ostream &out, bignum &c)
{
    out<<c.print();
    return out;
}

istream &operator >> (istream &in,bignum &c)
{
    char s[10000];
    in>>s;
    c = s;
    return in;
}
bignum gcd(bignum a,bignum b)
{
    return b==0?a:gcd(b,a%b);
}
int main()
{
    bignum a[255];
    a[0]=1;
    a[1]=1;
    a[2]=3;
    int n,i;
    for(i=3;i<=250;i++)
    {
        a[i]=a[i-1]+a[i-2]+a[i-2];
    }
    while(cin>>n)
    {
        cout<<a[n]<<endl;
    }
}

今天新加入的Java代码：

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

public class Main
{
	public static void main(String[] args)
	{

		// TODO Auto-generated method stub
		/*sss*/
		BigInteger [] a=new BigInteger[255]; //大数数组
		a[0]=BigInteger.ONE; //初始化
		a[1]=BigInteger.ONE;
		a[2]=a[1].add(a[0].add(a[0]));
		for(int i=3;i<=250;i++)
		{
			a[i]=a[i-1].add(a[i-2].add(a[i-2]));
		}
		Scanner cin = new Scanner(System.in); //输入打开
		
		while(cin.hasNext())//判断是否文件结束
		{
			int n=cin.nextInt();
			System.out.println(a[n]); //输出
		}
		cin .close();//输入关闭
	}
}