maximumSub-sequenceProduct3.cpp

/*
===========================================================
 * @名称:   UVA-787 - Maximum Sub-sequence Product	
 * @作者:		梁珞圣 
 * @创建时间: 2018-04-22 19:20:49 
 * @修改人:   梁珞圣 
 * @修改时间: 2018-04-22 19:20:49 
 * @备注:		评测时间0.000s
 * @题目来源： http://uva.onlinejudge.org/external/7/787.pdf
 * @思路:	
 * 1.结果超出long long，需要使用C++大整数；
 * 2.动态规划；
 * 3.因为乘数有负数，所以需要保留最小值（
 最小的负数乘以负数就是最大值）；
 * 4.同时用数组存储以i为结束的子段最大乘积，和最小乘积；
 * 5.建立递推方程，处理关系，参见maximumSub-sequenceProduct3.xlsx：
 *  1）以i为结束的子段最大乘积max(i)
 *  2）以i为结束的子段最小乘积min(i)
 *  3）以i-1为结束的子段最大乘积max(i-1) 
 *  4）以i-1为结束的子段最小乘积min(i-1)
 *  5）第i个值item(i)
 *  6) 不可能成立 min(i) > max（i）,排除不考虑；
    +---------------------------+----------+---+----------------------+---+----------------------+
    |                           | item(i)  | A |          >0          | 0 |          <0          |
    +---------------------------+----------+---+----------------------+---+----------------------+
    | 以i-1为结束的子段最大乘积     | max(i-1) | B |     >0    |    <0    |   |     >0    |    <0    |
    +---------------------------+----------+---+-----------+----------+---+-----------+----------+
    | 以i-1为结束的子段最小乘积     | min(i-1) | C | >0  | <0  | >0 | <0  |   | >0  | <0  | >0 | <0  |
    +---------------------------+----------+---+-----+-----+----+-----+---+-----+-----+----+-----+
    | 以i为结束的子段最大乘积       | max(i)   | D | B*A | B*C | -  | A   | 0 | A   | C*A | -  | C*A |
    +---------------------------+----------+---+-----+-----+----+-----+---+-----+-----+----+-----+
    | 以i为结束的子段最小乘积       | min(i)   | E | A   | C*A | -  | C*A | 0 | C*A | B*A | -  | A   |
    +---------------------------+----------+---+-----+-----+----+-----+---+-----+-----+----+-----+
 * 
===========================================================
*/
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <fstream>
#include <climits>
#include <functional>
#include <iomanip>
#include <iostream>
#include <set>
#include <string>
#include <vector>

using namespace std;

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint
{
    vector<int> z;
    int sign;

    bigint() : sign(1)
    {
    }

    bigint(long long v)
    {
        *this = v;
    }

    bigint(const string &s)
    {
        read(s);
    }

    void operator=(const bigint &v)
    {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v)
    {
        sign = 1;
        if (v < 0)
            sign = -1, v = -v;
        z.clear();
        for (; v > 0; v = v / base)
            z.push_back(v % base);
    }

    bigint operator+(const bigint &v) const
    {
        if (sign == v.sign)
        {
            bigint res = v;

            for (int i = 0, carry = 0; i < (int)max(z.size(), v.z.size()) || carry; ++i)
            {
                if (i == (int)res.z.size())
                    res.z.push_back(0);
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if (carry)
                    res.z[i] -= base;
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint &v) const
    {
        if (sign == v.sign)
        {
            if (abs() >= v.abs())
            {
                bigint res = *this;
                for (int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i)
                {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if (carry)
                        res.z[i] += base;
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v)
    {
        if (v < 0)
            sign = -sign, v = -v;
        for (int i = 0, carry = 0; i < (int)z.size() || carry; ++i)
        {
            if (i == (int)z.size())
                z.push_back(0);
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            //asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur), "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const
    {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint &a1, const bigint &b1)
    {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for (int i = a.z.size() - 1; i >= 0; i--)
        {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while (r < 0)
                r += b, --d;
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint &a1)
    {
        bigint a = a1;
        while (a.z.empty() || a.z.size() % 2 == 1)
            a.z.push_back(0);

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while (a.z.empty() || a.z.size() % 2 == 1)
            a.z.push_back(0);

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for (int j = n / 2 - 1; j >= 0; j--)
        {
            for (;; --q)
            {
                bigint r1 = (r - (res * 2 * base + q) * q) * base * base + (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
                if (r1 >= 0)
                {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if (j > 0)
            {
                int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) / (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint &v) const
    {
        return divmod(*this, v).first;
    }

    bigint operator%(const bigint &v) const
    {
        return divmod(*this, v).second;
    }

    void operator/=(int v)
    {
        if (v < 0)
            sign = -sign, v = -v;
        for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i)
        {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const
    {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const
    {
        if (v < 0)
            v = -v;
        int m = 0;
        for (int i = z.size() - 1; i >= 0; --i)
            m = (z[i] + m * (long long)base) % v;
        return m * sign;
    }

    void operator+=(const bigint &v)
    {
        *this = *this + v;
    }
    void operator-=(const bigint &v)
    {
        *this = *this - v;
    }
    void operator*=(const bigint &v)
    {
        *this = *this * v;
    }
    void operator/=(const bigint &v)
    {
        *this = *this / v;
    }

    bool operator<(const bigint &v) const
    {
        if (sign != v.sign)
            return sign < v.sign;
        if (z.size() != v.z.size())
            return z.size() * sign < v.z.size() * v.sign;
        for (int i = z.size() - 1; i >= 0; i--)
            if (z[i] != v.z[i])
                return z[i] * sign < v.z[i] * sign;
        return false;
    }

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

    void trim()
    {
        while (!z.empty() && z.back() == 0)
            z.pop_back();
        if (z.empty())
            sign = 1;
    }

    bool isZero() const
    {
        return z.empty() || (z.size() == 1 && !z[0]);
    }

    bigint operator-() const
    {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const
    {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const
    {
        long long res = 0;
        for (int i = z.size() - 1; i >= 0; i--)
            res = res * base + z[i];
        return res * sign;
    }

    friend bigint gcd(const bigint &a, const bigint &b)
    {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint &a, const bigint &b)
    {
        return a / gcd(a, b) * b;
    }

    void read(const string &s)
    {
        sign = 1;
        z.clear();
        int pos = 0;
        while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+'))
        {
            if (s[pos] == '-')
                sign = -sign;
            ++pos;
        }
        for (int i = s.size() - 1; i >= pos; i -= base_digits)
        {
            int x = 0;
            for (int j = max(pos, i - base_digits + 1); j <= i; j++)
                x = x * 10 + s[j] - '0';
            z.push_back(x);
        }
        trim();
    }

    friend istream &operator>>(istream &stream, bigint &v)
    {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream &operator<<(ostream &stream, const bigint &v)
    {
        if (v.sign == -1)
            stream << '-';
        stream << (v.z.empty() ? 0 : v.z.back());
        for (int i = (int)v.z.size() - 2; i >= 0; --i)
            stream << setw(base_digits) << setfill('0') << v.z[i];
        return stream;
    }

    static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits)
    {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for (int i = 1; i < (int)p.size(); i++)
            p[i] = p[i - 1] * 10;
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for (int i = 0; i < (int)a.size(); i++)
        {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while (cur_digits >= new_digits)
            {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while (!res.empty() && res.back() == 0)
            res.pop_back();
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll &a, const vll &b)
    {
        int n = a.size();
        vll res(n + n);
        if (n <= 32)
        {
            for (int i = 0; i < n; i++)
                for (int j = 0; j < n; j++)
                    res[i + j] += a[i] * b[j];
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for (int i = 0; i < k; i++)
            a2[i] += a1[i];
        for (int i = 0; i < k; i++)
            b2[i] += b1[i];

        vll r = karatsubaMultiply(a2, b2);
        for (int i = 0; i < (int)a1b1.size(); i++)
            r[i] -= a1b1[i];
        for (int i = 0; i < (int)a2b2.size(); i++)
            r[i] -= a2b2[i];

        for (int i = 0; i < (int)r.size(); i++)
            res[i + k] += r[i];
        for (int i = 0; i < (int)a1b1.size(); i++)
            res[i] += a1b1[i];
        for (int i = 0; i < (int)a2b2.size(); i++)
            res[i + n] += a2b2[i];
        return res;
    }

    bigint operator*(const bigint &v) const
    {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while (a.size() < b.size())
            a.push_back(0);
        while (b.size() < a.size())
            b.push_back(0);
        while (a.size() & (a.size() - 1))
            a.push_back(0), b.push_back(0);
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for (int i = 0, carry = 0; i < (int)c.size(); i++)
        {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

void solve()
{
    ifstream fin("maximumSub-sequenceProduct.in");
    ofstream fout("maximumSub-sequenceProduct.out");    

    while (true)
    {
        vector<int> inputData;  //输入的序列

        while (true)
        {
            int in = -1000000;
            fin >> in;

            if (in == -1000000)
            {
                return;
            }
            else if (in == -999999)
            {
                break;
            }
            inputData.push_back(in);
        }

        int size = inputData.size();

        if(size>0)
        {
            vector<bigint> maxProduct;  //使用数组存储：以i为结束的子段最大乘积
            vector<bigint> minProduct;  //使用数组存储：以i为结束的子段最小乘积

            //maxProduct初始值
            maxProduct.push_back(inputData[0]);
            
            //minProduct初始值
            minProduct.push_back(inputData[0]);

            for(int i=1;i<=size-1;++i)
            {
                int currItem=inputData[i];

                bigint iMax=0;      //以i为结束的子段最大乘积
                bigint iMin=0;      //以i为结束的子段最小乘积

                if(currItem>0)
                {
                    if(maxProduct[i-1]>0 && minProduct[i-1]>0)
                    {
                        iMax = maxProduct[i-1] * currItem;
                        iMin = currItem;
                    }
                    else if(maxProduct[i-1]>0 && minProduct[i-1]<0)
                    {
                        iMax = maxProduct[i-1] * currItem;
                        iMin = minProduct[i-1] * currItem;
                    }
                    else if(maxProduct[i-1]<0 && minProduct[i-1]<0)
                    {
                        iMax = currItem;
                        iMin = minProduct[i-1] * currItem;
                    }
                    else if(maxProduct[i-1]==0 && minProduct[i-1]==0)
                    {
                        iMax = currItem;
                        iMin = currItem;
                    }                    
                }
                else if(currItem<0)
                {
                    if(maxProduct[i-1]>0 && minProduct[i-1]>0)
                    {
                        iMax = currItem;
                        iMin = maxProduct[i-1] * currItem;
                    }
                    else if(maxProduct[i-1]>0 && minProduct[i-1]<0)
                    {
                        iMax = minProduct[i-1] * currItem;
                        iMin = maxProduct[i-1] * currItem;
                    }
                    else if(maxProduct[i-1]<0 && minProduct[i-1]<0)
                    {
                        iMax = minProduct[i-1] * currItem;
                        iMin = currItem;
                    }
                    else if(maxProduct[i-1]==0 && minProduct[i-1]==0)
                    {
                        iMax = currItem;
                        iMin = currItem;
                    }
                }

                maxProduct.push_back(iMax);
                minProduct.push_back(iMin);
            }

            //取出数组中的最大值就是答案
            auto it= max_element(maxProduct.begin(),maxProduct.end());
            fout << *it << '\n';
        }
        else
        {
            fout << 0 << '\n';
        }
        
    }
}

int main()
{
    solve();
    return 0;
}