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P1655 小朋友的球 题解

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c++ 题解

题目描述

横着看是一个斯特林数的问题，很容易想出方程：

S(n,m)=S(n−1,m−1)+S(n−1,m)∗mS(n,m) = S(n-1,m-1) + S(n-1,m)*m

下面给出证(xia)明(cou)过程：

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对于S(n,m)集合，它包括以下几个部分：
1. 从S(n-1,m-1)转移过来，即：最后一个球单独放一个盒子。
2. 从S(n-1,m)转移过来，即：它和某个共处，共有m种

此题唯一难点：
需要高精
参考代码：

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#include <bits/stdc++.h>
using namespace std;
class BigInt
{
#define Value(x, nega) ((nega) ? -(x) : (x))
#define At(vec, index) ((index) < vec.size() ? vec[(index)] : 0)
    //C风格的比较函数,其正负等于abs(lhs)-abs(rhs)的正负
    static int absComp(const BigInt &lhs, const BigInt &rhs)
    {
        if (lhs.size() != rhs.size())
            return lhs.size() < rhs.size() ? -1 : 1;
        for (int i = lhs.size() - 1; i >= 0; --i)
            if (lhs[i] != rhs[i])
                return lhs[i] < rhs[i] ? -1 : 1;
        return 0;
    }
    using Long = long long;
    const static int Exp = 9;
    const static Long Mod = 1000000000;
    mutable std::vector<Long> val;
    mutable bool nega = false;
    //规定:0的nega必须是false,0的size必须是0
    void trim() const
    {
        while (val.size() && val.back() == 0)
            val.pop_back();
        if (val.empty())
            nega = false;
    }
    int size() const { return val.size(); }
    Long &operator[](int index) const { return val[index]; }
    Long &back() const { return val.back(); }
    BigInt(int size, bool nega) : val(size), nega(nega) {}
    BigInt(const std::vector<Long> &val, bool nega) : val(val), nega(nega) {}

  public:
    friend std::ostream &operator<<(std::ostream &os, const BigInt &n)
    {
        if (n.size())
        {
            if (n.nega)
                putchar('-');
            for (int i = n.size() - 1; i >= 0; --i)
            {
                if (i == n.size() - 1)
                    printf("%lld", n[i]);
                else
                    printf("%0*lld", n.Exp, n[i]);
            }
        }
        else
            putchar('0');
        return os;
    }
    friend BigInt operator+(const BigInt &lhs, const BigInt &rhs)
    {
        BigInt ret(lhs);
        return ret += rhs;
    }
    friend BigInt operator-(const BigInt &lhs, const BigInt &rhs)
    {
        BigInt ret(lhs);
        return ret -= rhs;
    }
    BigInt(Long x = 0)
    {
        if (x < 0)
            x = -x, nega = true;
        while (x >= Mod)
            val.push_back(x % Mod), x /= Mod;
        if (x)
            val.push_back(x);
    }
    BigInt(const char *s)
    {
        int bound = 0, pos;
        if (s[0] == '-')
            nega = true, bound = 1;
        Long cur = 0, pow = 1;
        for (pos = strlen(s) - 1; pos >= Exp + bound - 1; pos -= Exp, val.push_back(cur), cur = 0, pow = 1)
            for (int i = pos; i > pos - Exp; --i)
                cur += (s[i] - '0') * pow, pow *= 10;
        for (cur = 0, pow = 1; pos >= bound; --pos)
            cur += (s[pos] - '0') * pow, pow *= 10;
        if (cur)
            val.push_back(cur);
    }
    BigInt &operator+=(const BigInt &rhs)
    {
        const int cap = std::max(size(), rhs.size()) + 1;
        val.resize(cap);
        int carry = 0;
        for (int i = 0; i < cap - 1; ++i)
        {
            val[i] = Value(val[i], nega) + Value(At(rhs, i), rhs.nega) + carry, carry = 0;
            if (val[i] >= Mod)
                val[i] -= Mod, carry = 1; //至多只需要减一次，不需要取模
            else if (val[i] < 0)
                val[i] += Mod, carry = -1; //同理
        }
        if ((val.back() = carry) == -1) //assert(val.back() == 1 or 0 or -1)
        {
            nega = true, val.pop_back();
            bool tailZero = true;
            for (int i = 0; i < cap - 1; ++i)
            {
                if (tailZero && val[i])
                    val[i] = Mod - val[i], tailZero = false;
                else
                    val[i] = Mod - 1 - val[i];
            }
        }
        trim();
        return *this;
    }
    friend BigInt operator-(const BigInt &rhs)
    {
        BigInt ret(rhs);
        ret.nega ^= 1;
        return ret;
    }
    BigInt &operator-=(const BigInt &rhs)
    {
        rhs.nega ^= 1;
        *this += rhs;
        rhs.nega ^= 1;
        return *this;
    }
    //高精*高精没办法原地操作，所以实现operator*
    //高精*低精可以原地操作，所以operator*=
    friend BigInt operator*(const BigInt &lhs, const BigInt &rhs)
    {
        const int cap = lhs.size() + rhs.size() + 1;
        BigInt ret(cap, lhs.nega ^ rhs.nega);
        //j < l.size(),i - j < rhs.size() => i - rhs.size() + 1 <= j
        for (int i = 0; i < cap - 1; ++i) // assert(0 <= ret[cap-1] < Mod)
            for (int j = std::max(i - rhs.size() + 1, 0), up = std::min(i + 1, lhs.size()); j < up; ++j)
            {
                ret[i] += lhs[j] * rhs[i - j];
                ret[i + 1] += ret[i] / Mod, ret[i] %= Mod;
            }
        ret.trim();
        return ret;
    }
    BigInt &operator*=(const BigInt &rhs) { return *this = *this * rhs; }
    friend BigInt operator/(const BigInt &lhs, const BigInt &rhs)
    {
        static std::vector<BigInt> powTwo{BigInt(1)};
        static std::vector<BigInt> estimate;
        estimate.clear();
        if (absComp(lhs, rhs) < 0)
            return BigInt();
        BigInt cur = rhs;
        int cmp;
        while ((cmp = absComp(cur, lhs)) <= 0)
        {
            estimate.push_back(cur), cur += cur;
            if (estimate.size() >= powTwo.size())
                powTwo.push_back(powTwo.back() + powTwo.back());
        }
        if (cmp == 0)
            return BigInt(powTwo.back().val, lhs.nega ^ rhs.nega);
        BigInt ret = powTwo[estimate.size() - 1];
        cur = estimate[estimate.size() - 1];
        for (int i = estimate.size() - 1; i >= 0 && cmp != 0; --i)
            if ((cmp = absComp(cur + estimate[i], lhs)) <= 0)
                cur += estimate[i], ret += powTwo[i];
        ret.nega = lhs.nega ^ rhs.nega;
        return ret;
    }
    bool operator==(const BigInt &rhs) const
    {
        return nega == rhs.nega && val == rhs.val;
    }
    bool operator!=(const BigInt &rhs) const { return nega != rhs.nega || val != rhs.val; }
    bool operator>=(const BigInt &rhs) const { return !(*this < rhs); }
    bool operator>(const BigInt &rhs) const { return !(*this <= rhs); }
    bool operator<=(const BigInt &rhs) const
    {
        if (nega && !rhs.nega)
            return true;
        if (!nega && rhs.nega)
            return false;
        int cmp = absComp(*this, rhs);
        return nega ? cmp >= 0 : cmp <= 0;
    }
    bool operator<(const BigInt &rhs) const
    {
        if (nega && !rhs.nega)
            return true;
        if (!nega && rhs.nega)
            return false;
        return (absComp(*this, rhs) < 0) ^ nega;
    }
    void swap(const BigInt &rhs) const
    {
        std::swap(val, rhs.val);
        std::swap(nega, rhs.nega);
    }
};
// const int N = 1e4 + 10;
// char a[N], b[N];
// int main()
// {
//     scanf("%s%s", a, b);
//     BigInt ba(a), bb(b);
//     std::cout << ba + bb << '\n';
//     std::cout << ba - bb << '\n';
//     std::cout << ba * bb << '\n';
//     BigInt d;
//     std::cout << (d = ba / bb) << '\n';
//     std::cout << ba - d * bb << '\n';
//     return 0;
// }
const int maxn = 1010;
BigInt dp[maxn][maxn];
BigInt f(int n,int m)//n球m
{
	if(dp[n][m] != BigInt())return dp[n][m];
	if(n < m || m <= 0)return BigInt();
	if(m == n)return 1;
	return dp[n][m] = f(n-1,m-1) + f(n-1,m) * m;
}
int main(void)
{
	int n,m;
	while(cin>>n>>m)
	{
		BigInt aa = f(n,m);
		cout<<aa<<endl;
	}
}

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