链接:https://ac.nowcoder.com/acm/contest/3005/E
来源:牛客网
 

时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 262144K,其他语言524288K
64bit IO Format: %lld

题目描述

给出一个包含数字1-9和加号的字符串,请你将字符串中的字符任意排列,但每种字符数目不变,使得结果是一个合法的表达式,而且表达式的值最小。输出那个最小表达式的值

合法的表达式的定义如下:

- 一个数字,如233,是一个合法的表达式

- A + B是合法的表达式,当且仅当 A , B 都是合法的表达式

保证给出的表达式经过重排,存在一个合法的解。

输入描述:

 

一行输入一个字符串,仅包含数字1-9和加号+。

字符串的长度小于5∗1055*10^55∗105。

输出描述:

一行输出一个数字,代表最小的解。

示例1

输入

 

111+1

输出

 

22

说明

11+11=22

示例2

输入

 

9998765432111

输出

 

1112345678999

示例3

输入

 

12+35

输出

 

38

说明

 

25+13 = 38

示例4

输入

 

23984692+238752+2+34+

输出

 

5461

说明

 

嗯,这个答案是可以得到的

备注:

注意,答案长度可能长达500000个字符。

题意:

给出一串数字和若干个加号,求所有元素重拍后能得到的最小的结果。

思路:

将所有数字从小到大排序,从较大数选取依次作为个位、十位、百位、、、

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 5e5 + 5;

int a[N], b[N];

int main()
{
    string s;
    while(getline(cin, s))
    {
        int len = s.size();
        int cn = 0;
        int tot = 0;
        for(int i = 0; i < s.size(); ++i)
        {
            if(s[i] == '+')
                cn++;
            else
                a[++tot] = s[i] - '0';
        }
        sort(a + 1, a + tot + 1);
        cn++;

        ll tmp = 0;
        int id = 0;
        for(int i = tot; i >= 1; --i)
        {
            tmp += a[i];
            if((tot - i + 1) % cn == 0)
            {
                b[++id] = tmp % 10;
                tmp /= 10;
            }
        }
        if(tmp)
        {
            b[++id] = tmp;
        }
        for(int i = id; i >= 1; --i)
        {
            cout<<b[i];
        }
        cout<<'\n';
    }
    return 0;
}

或者采用大数类:

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 5e5 + 5;

//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;
    }
};

int main()
{
    string s;
    getline(cin, s);
    int len = s.size();
    sort(s.begin(), s.end());
    int cn = 0;
    for(int i = 0; i < len; ++i)
    {
        if(s[i] == '+')
            cn++;
        else
            break;
    }
    BigInteger ans = 0;
    string tmp;
    for(int i = cn; i < 2 * cn + 1; ++i)
    {
        tmp = "";
        for(int j = i; j < len; j += (cn + 1))
            tmp += s[j];
        ans += BigInteger(tmp);
    }
    cout<<ans<<'\n';
    return 0;
}