洛谷P1018 乘积最大

洛谷P1018 乘积最大

题意：

给一个长度为n的数字串，在这个数字串中插入k个乘号，使得表达式的乘积最大

分析一下：

算了，懒得分析了，代码中有详细注释，直接看代码吧…

代码君：

#include 
using namespace std;
const int maxn = 45;

struct BigInteger
{
    typedef unsigned long long LL;

    static const int BASE = 100000000;
    static const int WIDTH = 8;
    vector<int> s;

    BigInteger &clean()
    {
        while (!s.back() && s.size() > 1)
            s.pop_back();
        return *this;
    }
    BigInteger(LL num = 0)
    {
        *this = num;
    }
    BigInteger(string s)
    {
        *this = s;
    }
    BigInteger &operator=(long long num)
    {
        s.clear();
        do
        {
            s.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }
    BigInteger &operator=(const string &str)
    {
        s.clear();
        int x, len = (str.length() - 1) / WIDTH + 1;
        for (int i = 0; i < len; i++)
        {
            int end = str.length() - i * WIDTH;
            int start = max(0, end - WIDTH);
            sscanf(str.substr(start, end - start).c_str(), "%d", &x);
            s.push_back(x);
        }
        return (*this).clean();
    }

    BigInteger operator+(const BigInteger &b) const
    {
        BigInteger c;
        c.s.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= s.size() && i >= b.s.size())
                break;
            int x = g;
            if (i < s.size())
                x += s[i];
            if (i < b.s.size())
                x += b.s[i];
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c;
    }
    BigInteger operator-(const BigInteger &b) const
    {
        assert(b <= *this); // 减数不能大于被减数
        BigInteger c;
        c.s.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= s.size() && i >= b.s.size())
                break;
            int x = s[i] + g;
            if (i < b.s.size())
                x -= b.s[i];
            if (x < 0)
            {
                g = -1;
                x += BASE;
            }
            else
                g = 0;
            c.s.push_back(x);
        }
        return c.clean();
    }
    BigInteger operator*(const BigInteger &b) const
    {
        int i, j;
        LL g;
        vector<LL> v(s.size() + b.s.size(), 0);
        BigInteger c;
        c.s.clear();
        for (i = 0; i < s.size(); i++)
            for (j = 0; j < b.s.size(); j++)
                v[i + j] += LL(s[i]) * b.s[j];
        for (i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= v.size())
                break;
            LL x = v[i] + g;
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c.clean();
    }
    BigInteger operator/(const BigInteger &b) const
    {
        assert(b > 0);        // 除数必须大于0
        BigInteger c = *this; // 商:主要是让c.s和(*this).s的vector一样大
        BigInteger m;         // 余数:初始化为0
        for (int i = s.size() - 1; i >= 0; i--)
        {
            m = m * BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b * c.s[i];
        }
        return c.clean();
    }
    BigInteger operator%(const BigInteger &b) const //方法与除法相同
    {
        BigInteger c = *this;
        BigInteger m;
        for (int i = s.size() - 1; i >= 0; i--)
        {
            m = m * BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b * c.s[i];
        }
        return m;
    }

    int bsearch(const BigInteger &b, const BigInteger &m) const
    {
        int L = 0, R = BASE - 1, x;
        while (1)
        {
            x = (L + R) >> 1;
            if (b * x <= m)
            {
                if (b * (x + 1) > m)
                    return x;
                else
                    L = x;
            }
            else
                R = x;
        }
    }
    BigInteger &operator+=(const BigInteger &b)
    {
        *this = *this + b;
        return *this;
    }
    BigInteger &operator-=(const BigInteger &b)
    {
        *this = *this - b;
        return *this;
    }
    BigInteger &operator*=(const BigInteger &b)
    {
        *this = *this * b;
        return *this;
    }
    BigInteger &operator/=(const BigInteger &b)
    {
        *this = *this / b;
        return *this;
    }
    BigInteger &operator%=(const BigInteger &b)
    {
        *this = *this % b;
        return *this;
    }

    bool operator<(const BigInteger &b) const
    {
        if (s.size() != b.s.size())
            return s.size() < b.s.size();
        for (int i = s.size() - 1; i >= 0; i--)
            if (s[i] != b.s[i])
                return s[i] < b.s[i];
        return false;
    }
    bool operator>(const BigInteger &b) const
    {
        return b < *this;
    }
    bool operator<=(const BigInteger &b) const
    {
        return !(b < *this);
    }
    bool operator>=(const BigInteger &b) const
    {
        return !(*this < b);
    }
    bool operator!=(const BigInteger &b) const
    {
        return b < *this || *this < b;
    }
    bool operator==(const BigInteger &b) const
    {
        return !(b < *this) && !(b > *this);
    }
};

ostream &operator<<(ostream &out, const BigInteger &x)
{
    out << x.s.back();
    for (int i = x.s.size() - 2; i >= 0; i--)
    {
        char buf[20];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = 0; j < strlen(buf); j++)
            out << buf[j];
    }
    return out;
}

istream &operator>>(istream &in, BigInteger &x)
{
    string s;
    if (!(in >> s))
        return in;
    x = s;
    return in;
}

inline BigInteger MAX(const BigInteger &a, const BigInteger &b)
{
    if (a > b)
        return a;
    return b;
}
// 以上全为高精度板子...

BigInteger dp[maxn][10], a[maxn][maxn];
//dp[i][j]表示前i位数用了j个乘号的最大值
//a[i][j]表示第i位到第j位数字连在一起组成的数

int main()
{
    int n, k;
    string s;
    cin >> n >> k;
    cin >> s;

    for (int i = 0; i < n; ++i)
        a[i + 1][i + 1] = s.substr(i, 1); //分离每一个数字，并存到数组a中

    for (int i = 2; i <= n; ++i) //预处理长度大于一的子串，将第j位到第i位数字连在一起组成的数存入a中
        for (int j = i - 1; j >= 1; --j)
        {
            a[j][i] = a[j][i - 1] * 10 + a[i][i];
        }

    for (int i = 1; i <= n; ++i)
        dp[i][0] = a[1][i]; //初始化dp数组

    for (int i = 1; i <= k; ++i)
        for (int j = i + 1; j <= n; ++j)
            for (int v = i; v < j; ++v)
            {
                dp[j][i] = MAX(dp[v][i - 1] * a[v + 1][j], dp[j][i]); //d[j][i]的最优值为dp[v][i-1]*a[v+1][j]中的最大值
            }
    cout << dp[n][k] << endl;
    return 0;
}