codeforces gym 102028 E Resistors in Parallel 大数+贪心_gym

本文链接：https://blog.youkuaiyun.com/xiji333/article/details/101111465

本文详细解析了Codeforces上一道关于电阻并联的数学问题，通过观察规律找到了一个高效的递推解决方案，使用了素数筛选算法和大数运算来处理等效电阻的计算。

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http://codeforces.com/gym/102028/problem/E
题目大意：给你 $n$ 个电阻，第 $i$ 个电阻的阻值为：(1)若 $i$ 能够被 $d^{2}$ 整除，其中( $d > = 2$ )则 $R_{i}=\inf$ ；(2)否则 $R_{i}=i$ 。你有 $n$ 种并联电阻的方式，第 $i$ 种方式是把下标为 $i$ 的因子的电阻并联起来，求并联后的等效电阻的最小值。(化成最简分数的形式)

思路：电阻并联的计算公式如下。
在这里插入图片描述
那么要使 $R$ 最小，就要使分母最大，那么就要使 $R_{1}、R_{2}、……、R_{n}$ 尽量小，再根据电阻阻值的定义，就知道最优选择就是从小到大的选取素数。但是阻值的计算不是很方便，但是打表很容易看出规律。举个例子：(1)当 $n = 1$ 时， $R_{1}=1/1=1$ ；(2)当 $n = 2$ 时， $R_{2}=1/(1+1/2)=2/3$ ；(3)当 $n = 6$ 时， $R_{3}=1/(1+1/2+1/3+1/6)=1/2$ ；(4)当 $n = 30$ 时，同理可计算出 $R_{4}=5/12$ 。可以发现以下规律： $R_{2}=R_{1}*2/3,\ R_{3}=R_{2}*3/4,\ R_{4}=R_{3}*5/6$ (分子是素数分母是素数+1)，我们得到了一个递推关系，那么就可以很快计算出答案辣。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define EPS 1e-10
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

const int maxn=305;
bool vis[maxn+5];
ll prime[maxn+5];
int k=-1;

void oula()
{
	for(int i=2;i<=maxn;i++)
	{
		if(!vis[i])
			prime[++k]=i;
		for(int j=0;j<=k&&prime[j]*i<=maxn;j++)
		{
			vis[prime[j]*i]=1;
			if(i%prime[j]==0)
				break;
		}
	}
}

//大数
struct BigInteger
{
    static const int BASE = 100000000; //和WIDTH保持一致
    static const int WIDTH = 8;        //八位一存储,如修改记得修改输出中的%08d
    bool sign;                         //符号, 0表示负数
    size_t length;                     //位数
    vector<int> num;                   //反序存
                                       //构造函数
    BigInteger(long long x = 0) { *this = x; }
    BigInteger(const string &x) { *this = x; }
    BigInteger(const BigInteger &x) { *this = x; }
    //剪掉前导0，并且求一下数的位数
    void cutLeadingZero()
    {
        while (num.back() == 0 && num.size() != 1)
        {
            num.pop_back();
        }
        int tmp = num.back();
        if (tmp == 0)
        {
            length = 1;
        }
        else
        {
            length = (num.size() - 1) * WIDTH;
            while (tmp > 0)
            {
                length++;
                tmp /= 10;
            }
        }
    }
    //赋值运算符
    BigInteger &operator=(long long x)
    {
        num.clear();
        if (x >= 0)
        {
            sign = true;
        }
        else
        {
            sign = false;
            x = -x;
        }
        do
        {
            num.push_back(x % BASE);
            x /= BASE;
        } while (x > 0);
        cutLeadingZero();
        return *this;
    }
    BigInteger &operator=(const string &str)
    {
        num.clear();
        sign = (str[0] != '-'); //设置符号
        int x, len = (str.size() - 1 - (!sign)) / WIDTH + 1;
        for (int i = 0; i < len; i++)
        {
            int End = str.size() - i * WIDTH;
            int start = max((int)(!sign), End - WIDTH); //防止越界
            sscanf(str.substr(start, End - start).c_str(), "%d", &x);
            num.push_back(x);
        }
        cutLeadingZero();
        return *this;
    }
    BigInteger &operator=(const BigInteger &tmp)
    {
        num = tmp.num;
        sign = tmp.sign;
        length = tmp.length;
        return *this;
    }
    //绝对值
    BigInteger abs() const
    {
        BigInteger ans(*this);
        ans.sign = true;
        return ans;
    }
    //正号
    const BigInteger &operator+() const { return *this; }
    //负号
    BigInteger operator-() const
    {
        BigInteger ans(*this);
        if (ans != 0)
            ans.sign = !ans.sign;
        return ans;
    }
    // + 运算符
    BigInteger operator+(const BigInteger &b) const
    {
        if (!b.sign)
        {
            return *this - (-b);
        }
        if (!sign)
        {
            return b - (-*this);
        }
        BigInteger ans;
        ans.num.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= num.size() && i >= b.num.size())
                break;
            int x = g;
            if (i < num.size())
                x += num[i];
            if (i < b.num.size())
                x += b.num[i];
            ans.num.push_back(x % BASE);
            g = x / BASE;
        }
        ans.cutLeadingZero();
        return ans;
    }
    // - 运算符
    BigInteger operator-(const BigInteger &b) const
    {
        if (!b.sign)
        {
            return *this + (-b);
        }
        if (!sign)
        {
            return -((-*this) + b);
        }
        if (*this < b)
        {
            return -(b - *this);
        }
        BigInteger ans;
        ans.num.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= num.size() && i >= b.num.size())
                break;
            int x = g;
            g = 0;
            if (i < num.size())
                x += num[i];
            if (i < b.num.size())
                x -= b.num[i];
            if (x < 0)
            {
                x += BASE;
                g = -1;
            }
            ans.num.push_back(x);
        }
        ans.cutLeadingZero();
        return ans;
    }
    // * 运算符
    BigInteger operator*(const BigInteger &b) const
    {
        int lena = num.size(), lenb = b.num.size();
        BigInteger ans;
        for (int i = 0; i < lena + lenb; i++)
            ans.num.push_back(0);
        for (int i = 0, g = 0; i < lena; i++)
        {
            g = 0;
            for (int j = 0; j < lenb; j++)
            {
                long long x = ans.num[i + j];
                x += (long long)num[i] * (long long)b.num[j];
                ans.num[i + j] = x % BASE;
                g = x / BASE;
                ans.num[i + j + 1] += g;
            }
        }
        ans.cutLeadingZero();
        ans.sign = (ans.length == 1 && ans.num[0] == 0) || (sign == b.sign);
        return ans;
    }
    //*10^n 大数除大数中用到
    BigInteger e(size_t n) const
    {
        int tmp = n % WIDTH;
        BigInteger ans;
        ans.length = n + 1;
        n /= WIDTH;
        while (ans.num.size() <= n)
            ans.num.push_back(0);
        ans.num[n] = 1;
        while (tmp--)
            ans.num[n] *= 10;
        return ans * (*this);
    }
    // /运算符 (大数除大数)
    BigInteger operator/(const BigInteger &b) const
    {
        BigInteger aa((*this).abs());
        BigInteger bb(b.abs());
        if (aa < bb)
            return 0;
        char *str = new char[aa.length + 1];
        memset(str, 0, sizeof(char) * (aa.length + 1));
        BigInteger tmp;
        int lena = aa.length, lenb = bb.length;
        for (int i = 0; i <= lena - lenb; i++)
        {
            tmp = bb.e(lena - lenb - i);
            while (aa >= tmp)
            {
                str[i]++;
                aa = aa - tmp;
            }
            str[i] += '0';
        }
        BigInteger ans(str);
        delete[] str;
        ans.sign = (ans == 0 || sign == b.sign);
        return ans;
    }
    // %运算符
    BigInteger operator%(const BigInteger &b) const
    {
        return *this - *this / b * b;
    }
    // ++ 运算符
    BigInteger &operator++()
    {
        *this = *this + 1;
        return *this;
    }
    // -- 运算符
    BigInteger &operator--()
    {
        *this = *this - 1;
        return *this;
    }
    // += 运算符
    BigInteger &operator+=(const BigInteger &b)
    {
        *this = *this + b;
        return *this;
    }
    // -= 运算符
    BigInteger &operator-=(const BigInteger &b)
    {
        *this = *this - b;
        return *this;
    }
    // *=运算符
    BigInteger &operator*=(const BigInteger &b)
    {
        *this = *this * b;
        return *this;
    }
    // /= 运算符
    BigInteger &operator/=(const BigInteger &b)
    {
        *this = *this / b;
        return *this;
    }
    // %=运算符
    BigInteger &operator%=(const BigInteger &b)
    {
        *this = *this % b;
        return *this;
    }
    // < 运算符
    bool operator<(const BigInteger &b) const
    {
        if (sign != b.sign) //正负，负正
        {
            return !sign;
        }
        else if (!sign && !b.sign) //负负
        {
            return -b < -*this;
        }
        //正正
        if (num.size() != b.num.size())
            return num.size() < b.num.size();
        for (int i = num.size() - 1; i >= 0; i--)
            if (num[i] != b.num[i])
                return num[i] < b.num[i];
        return false;
    }
    bool operator>(const BigInteger &b) const { return b < *this; }                     // >  运算符
    bool operator<=(const BigInteger &b) const { return !(b < *this); }                 // <= 运算符
    bool operator>=(const BigInteger &b) const { return !(*this < b); }                 // >= 运算符
    bool operator!=(const BigInteger &b) const { return b < *this || *this < b; }       // != 运算符
    bool operator==(const BigInteger &b) const { return !(b < *this) && !(*this < b); } //==运算符
    // 逻辑运算符
    bool operator||(const BigInteger &b) const { return *this != 0 || b != 0; } // || 运算符
    bool operator&&(const BigInteger &b) const { return *this != 0 && b != 0; } // && 运算符
    bool operator!() { return (bool)(*this == 0); }                             // ！ 运算符

    //重载<<使得可以直接输出大数
    friend ostream &operator<<(ostream &out, const BigInteger &x)
    {
        if (!x.sign)
            out << '-';
        out << x.num.back();
        for (int i = x.num.size() - 2; i >= 0; i--)
        {
            char buf[10];
            //如WIDTH和BASR有变化,此处要修改为%0(WIDTH)d
            sprintf(buf, "%08d", x.num[i]);
            for (int j = 0; j < strlen(buf); j++)
                out << buf[j];
        }
        return out;
    }
    //重载>>使得可以直接输入大数
    friend istream &operator>>(istream &in, BigInteger &x)
    {
        string str;
        in >> str;
        size_t len = str.size();
        int start = 0;
        if (str[0] == '-')
            start = 1;
        if (str[start] == '\0')
            return in;
        for (int i = start; i < len; i++)
        {
            if (str[i] < '0' || str[i] > '9')
                return in;
        }
        x.sign = !start;
        x = str.c_str();
        return in;
    }
};

BigInteger zero(0ll);
BigInteger n;

BigInteger gcd(BigInteger a,BigInteger b)
{
	return b==zero?a:gcd(b,a%b);
}

int main()
{
	oula();
	int t;
	scanf("%d",&t);
	while(t--)
	{
		cin>>n;
		BigInteger fz(1ll),fm(1ll);
		int i=0;
		while(fz*BigInteger(prime[i])<=n)
		{
			fz*=BigInteger(prime[i]);
			fm*=BigInteger(prime[i]+1);
			++i;
		}
		BigInteger tmp=gcd(fz,fm);
		fz/=tmp,fm/=tmp;
		cout<<fz<<"/"<<fm<<endl;
	}
    return 0;
}