按字典顺序排序C++的整数数组

Question

我想按字典顺序对大量整数(比如说1个元素)进行排序.

例:

input [] = { 100, 21 , 22 , 99 , 1  , 927 }
sorted[] = { 1  , 100, 21 , 22 , 927, 99  }

我用最简单的方法完成了它:

• 将所有数字转换为字符串(非常昂贵,因为它将占用大量内存)
• 使用std:sort与strcmp作为比较功能
• 将字符串转换回整数

有没有比这更好的方法？

Answer 1

使用std::sort()合适的比较功能.这减少了内存需求.

比较功能可以使用n % 10,n / 10 % 10,n / 100 % 10等来访问各个数字(正整数;负整数工作有点不同).

Answer 2

要提供任何自定义排序顺序,您可以提供比较器std::sort.在这种情况下,它会有些复杂,使用对数来检查基数为10的个数.

这是一个例子 - 评论内联描述了正在发生的事情.

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cassert>

int main() {
    int input[] { 100, 21, 22, 99, 1, 927, -50, -24, -160 };

    /**
     * Sorts the array lexicographically.
     * 
     * The trick is that we have to compare digits left-to-right
     * (considering typical Latin decimal notation) and that each of
     * two numbers to compare may have a different number of digits.
     * 
     * This is very efficient in storage space, but inefficient in
     * execution time; an approach that pre-visits each element and
     * stores a translated representation will at least double your
     * storage requirements (possibly a problem with large inputs)
     * but require only a single translation of each element.
     */
    std::sort(
        std::begin(input),
        std::end(input),
        [](int lhs, int rhs) -> bool {
            // Returns true if lhs < rhs
            // Returns false otherwise
            const auto BASE      = 10;
            const bool LHS_FIRST = true;
            const bool RHS_FIRST = false;
            const bool EQUAL     = false;


            // There's no point in doing anything at all
            // if both inputs are the same; strict-weak
            // ordering requires that we return `false`
            // in this case.
            if (lhs == rhs) {
                return EQUAL;
            }

            // Compensate for sign
            if (lhs < 0 && rhs < 0) {
                // When both are negative, sign on its own yields
                // no clear ordering between the two arguments.
                // 
                // Remove the sign and continue as for positive
                // numbers.
                lhs *= -1;
                rhs *= -1;
            }
            else if (lhs < 0) {
                // When the LHS is negative but the RHS is not,
                // consider the LHS "first" always as we wish to
                // prioritise the leading '-'.
                return LHS_FIRST;
            }
            else if (rhs < 0) {
                // When the RHS is negative but the LHS is not,
                // consider the RHS "first" always as we wish to
                // prioritise the leading '-'.
                return RHS_FIRST;
            }

            // Counting the number of digits in both the LHS and RHS
            // arguments is *almost* trivial.
            const auto lhs_digits = (
                lhs == 0
                ? 1
                : std::ceil(std::log(lhs+1)/std::log(BASE))
            );

            const auto rhs_digits = (
                rhs == 0
                ? 1
                : std::ceil(std::log(rhs+1)/std::log(BASE))
            );

            // Now we loop through the positions, left-to-right,
            // calculating the digit at these positions for each
            // input, and comparing them numerically. The
            // lexicographic nature of the sorting comes from the
            // fact that we are doing this per-digit comparison
            // rather than considering the input value as a whole.
            const auto max_pos = std::max(lhs_digits, rhs_digits);
            for (auto pos = 0; pos < max_pos; pos++) {
                if (lhs_digits - pos == 0) {
                    // Ran out of digits on the LHS;
                    // prioritise the shorter input
                    return LHS_FIRST;
                }
                else if (rhs_digits - pos == 0) {
                    // Ran out of digits on the RHS;
                    // prioritise the shorter input
                    return RHS_FIRST;
                }
                else {
                    const auto lhs_x = (lhs / static_cast<decltype(BASE)>(std::pow(BASE, lhs_digits - 1 - pos))) % BASE;
                    const auto rhs_x = (rhs / static_cast<decltype(BASE)>(std::pow(BASE, rhs_digits - 1 - pos))) % BASE;

                    if (lhs_x < rhs_x)
                        return LHS_FIRST;
                    else if (rhs_x < lhs_x)
                        return RHS_FIRST;
                }
            }

            // If we reached the end and everything still
            // matches up, then something probably went wrong
            // as I'd have expected to catch this in the tests
            // for equality.
            assert("Unknown case encountered");
        }
    );

    std::cout << '{';
    for (auto x : input)
        std::cout << x << ", ";
    std::cout << '}';

    // Output: {-160, -24, -50, 1, 100, 21, 22, 927, 99, }
}

演示

有更快的方法来计算数字中的位数,但上面的内容将帮助您入门.

Answer 3

这是另一种在排序之前完成一些计算的算法.尽管有额外的复制(见比较),它似乎相当快.

注意:

• 它只支持正整数
• in仅支持整数<= std::numeric_limits<int>::max()/10

_{NB你可以优化count_digits和my_pow10; 例如,请参阅Andrei Alexandrescu 撰写的三个C++优化技巧,以及比pow()更快地计算C++中10的整数幂？}

助手:

#include <random>
#include <vector>
#include <utility>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <limits>
#include <iterator>

// non-optimized version
int count_digits(int p) // returns `0` for `p == 0`
{
    int res = 0;
    for(; p != 0; ++res)
    {
        p /= 10;
    }
    return res;
}

// non-optimized version
int my_pow10(unsigned exp)
{
    int res = 1;
    for(; exp != 0; --exp)
    {
        res *= 10;
    }
    return res;
}

算法(注意 - 不是就地):

// helper to provide integers with the same number of digits
template<class T, class U>
std::pair<T, T> lexicographic_pair_helper(T const p, U const maxDigits)
{
    auto const digits = count_digits(p);
    // append zeros so that `l` has `maxDigits` digits
    auto const l = static_cast<T>( p  * my_pow10(maxDigits-digits) );
    return {l, p};
}

template<class RaIt>
using pair_vec
    = std::vector<std::pair<typename std::iterator_traits<RaIt>::value_type,
                            typename std::iterator_traits<RaIt>::value_type>>;

template<class RaIt>
pair_vec<RaIt> lexicographic_sort(RaIt p_beg, RaIt p_end)
{
    if(p_beg == p_end) return {};

    auto max = *std::max_element(p_beg, p_end);
    auto maxDigits = count_digits(max);

    pair_vec<RaIt> result;
    result.reserve( std::distance(p_beg, p_end) );

    for(auto i = p_beg; i != p_end; ++i)
        result.push_back( lexicographic_pair_helper(*i, maxDigits) );

    using value_type = typename pair_vec<RaIt>::value_type;

    std::sort(begin(result), end(result),
              [](value_type const& l, value_type const& r)
              {
                  if(l.first < r.first) return true;
                  if(l.first > r.first) return false;
                  return l.second < r.second; }
             );

    return result;
}

用法示例:

int main()
{
    std::vector<int> input = { 100, 21 , 22 , 99 , 1  , 927 };
    // generate some numbers
    /*{
        constexpr int number_of_elements = 1E6;
        std::random_device rd;
        std::mt19937 gen( rd() );
        std::uniform_int_distribution<>
            dist(0, std::numeric_limits<int>::max()/10);
        for(int i = 0; i < number_of_elements; ++i)
            input.push_back( dist(gen) );
    }*/

    std::cout << "unsorted: ";
    for(auto const& e : input) std::cout << e << ", ";
    std::cout << "\n\n";


    auto sorted = lexicographic_sort(begin(input), end(input));

    std::cout << "sorted: ";
    for(auto const& e : sorted) std::cout << e.second << ", ";
    std::cout << "\n\n";
}

Answer 4

这是一个社区维基,用于比较解决方案.我使用了nim的代码并使其易于扩展.随意添加您的解决方案和输出.

示例使用g ++ 4.9 @运行旧的慢速计算机(3 GB RAM,Core2Duo U9400)-O3 -march=native:

number of elements: 1e+03
size of integer type: 4

reference solution: Lightness Races in Orbit

solution "dyp":
    duration: 0 ms and 301 microseconds
    comparison to reference solution: exact match
solution "Nim":
    duration: 2 ms and 160 microseconds
    comparison to reference solution: exact match
solution "nyarlathotep":
    duration: 8 ms and 126 microseconds
    comparison to reference solution: exact match
solution "notbad":
    duration: 1 ms and 102 microseconds
    comparison to reference solution: exact match
solution "Eric Postpischil":
    duration: 2 ms and 550 microseconds
    comparison to reference solution: exact match
solution "Lightness Races in Orbit":
    duration: 17 ms and 469 microseconds
    comparison to reference solution: exact match
solution "pts":
    duration: 1 ms and 92 microseconds
    comparison to reference solution: exact match

==========================================================

number of elements: 1e+04
size of integer type: 4

reference solution: Lightness Races in Orbit

solution "nyarlathotep":
    duration: 109 ms and 712 microseconds
    comparison to reference solution: exact match
solution "Lightness Races in Orbit":
    duration: 272 ms and 819 microseconds
    comparison to reference solution: exact match
solution "dyp":
    duration: 1 ms and 748 microseconds
    comparison to reference solution: exact match
solution "notbad":
    duration: 16 ms and 115 microseconds
    comparison to reference solution: exact match
solution "pts":
    duration: 15 ms and 10 microseconds
    comparison to reference solution: exact match
solution "Eric Postpischil":
    duration: 33 ms and 301 microseconds
    comparison to reference solution: exact match
solution "Nim":
    duration: 17 ms and 83 microseconds
    comparison to reference solution: exact match

==========================================================

number of elements: 1e+05
size of integer type: 4

reference solution: Lightness Races in Orbit

solution "Nim":
    duration: 217 ms and 4 microseconds
    comparison to reference solution: exact match
solution "pts":
    duration: 199 ms and 505 microseconds
    comparison to reference solution: exact match
solution "dyp":
    duration: 20 ms and 330 microseconds
    comparison to reference solution: exact match
solution "Eric Postpischil":
    duration: 415 ms and 477 microseconds
    comparison to reference solution: exact match
solution "Lightness Races in Orbit":
    duration: 3955 ms and 58 microseconds
    comparison to reference solution: exact match
solution "notbad":
    duration: 215 ms and 259 microseconds
    comparison to reference solution: exact match
solution "nyarlathotep":
    duration: 1341 ms and 46 microseconds
    comparison to reference solution: mismatch found

==========================================================

number of elements: 1e+06
size of integer type: 4

reference solution: Lightness Races in Orbit

solution "Lightness Races in Orbit":
    duration: 52861 ms and 314 microseconds
    comparison to reference solution: exact match
solution "Eric Postpischil":
    duration: 4757 ms and 608 microseconds
    comparison to reference solution: exact match
solution "nyarlathotep":
    duration: 15654 ms and 195 microseconds
    comparison to reference solution: mismatch found
solution "dyp":
    duration: 233 ms and 779 microseconds
    comparison to reference solution: exact match
solution "pts":
    duration: 2181 ms and 634 microseconds
    comparison to reference solution: exact match
solution "Nim":
    duration: 2539 ms and 9 microseconds
    comparison to reference solution: exact match
solution "notbad":
    duration: 2675 ms and 362 microseconds
    comparison to reference solution: exact match

==========================================================

number of elements: 1e+07
size of integer type: 4

reference solution: Lightness Races in Orbit

solution "notbad":
    duration: 33425 ms and 423 microseconds
    comparison to reference solution: exact match
solution "pts":
    duration: 26000 ms and 398 microseconds
    comparison to reference solution: exact match
solution "Eric Postpischil":
    duration: 56206 ms and 359 microseconds
    comparison to reference solution: exact match
solution "Lightness Races in Orbit":
    duration: 658540 ms and 342 microseconds
    comparison to reference solution: exact match
solution "nyarlathotep":
    duration: 187064 ms and 518 microseconds
    comparison to reference solution: mismatch found
solution "Nim":
    duration: 30519 ms and 227 microseconds
    comparison to reference solution: exact match
solution "dyp":
    duration: 2624 ms and 644 microseconds
    comparison to reference solution: exact match

算法必须是具有支持接口的函数调用操作符模板的结构:

template<class RaIt> operator()(RaIt begin, RaIt end);

提供输入数据的副本作为参数,期望算法在相同范围内提供结果(例如,就地排序).

#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <random>
#include <vector>
#include <utility>
#include <cmath>
#include <cassert>
#include <chrono>
#include <cstring>
#include <climits>
#include <functional>
#include <cstdlib>
#include <iomanip>

using duration_t = decltype(  std::chrono::high_resolution_clock::now()
                            - std::chrono::high_resolution_clock::now());

template<class T>
struct result_t
{
    std::vector<T> numbers;
    duration_t duration;
    char const* name;
};

template<class RaIt, class F>
result_t<typename std::iterator_traits<RaIt>::value_type>
apply_algorithm(RaIt p_beg, RaIt p_end, F f, char const* name)
{
    using value_type = typename std::iterator_traits<RaIt>::value_type;

    std::vector<value_type> inplace(p_beg, p_end);

    auto start = std::chrono::high_resolution_clock::now();

    f(begin(inplace), end(inplace));

    auto end = std::chrono::high_resolution_clock::now();
    auto duration = end - start;

    return {std::move(inplace), duration, name};
}

// non-optimized version
int count_digits(int p) // returns `0` for `p == 0`
{
        int res = 0;
        for(; p != 0; ++res)
        {
            p /= 10;
        }
        return res;
}

// non-optimized version
int my_pow10(unsigned exp)
{
        int res = 1;
        for(; exp != 0; --exp)
        {
            res *= 10;
        }
        return res;
}


// !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
// paste algorithms here
// !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


int main(int argc, char** argv)
{
    using integer_t = int;
    constexpr integer_t dist_min = 0;
    constexpr integer_t dist_max = std::numeric_limits<integer_t>::max()/10;
    constexpr std::size_t default_number_of_elements = 1E6;

    const std::size_t number_of_elements = argc>1 ? std::atoll(argv[1]) :
                                           default_number_of_elements;
    std::cout << "number of elements: ";
    std::cout << std::scientific << std::setprecision(0);
    std::cout << (double)number_of_elements << "\n";
    std::cout << /*std::defaultfloat <<*/ std::setprecision(6);
    std::cout.unsetf(std::ios_base::floatfield);

    std::cout << "size of integer type: " << sizeof(integer_t) << "\n\n";

    std::vector<integer_t> input;
    {
        input.reserve(number_of_elements);

        std::random_device rd;
        std::mt19937 gen( rd() );
        std::uniform_int_distribution<> dist(dist_min, dist_max);

        for(std::size_t i = 0; i < number_of_elements; ++i)
            input.push_back( dist(gen) );
    }

    auto b = begin(input);
    auto e = end(input);

    using res_t = result_t<integer_t>;
    std::vector< std::function<res_t()> > algorithms;

    #define MAKE_BINDER(B, E, ALGO, NAME) \
        std::bind( &apply_algorithm<decltype(B),decltype(ALGO)>, \
                   B,E,ALGO,NAME )
    constexpr auto lightness_name = "Lightness Races in Orbit";
    algorithms.push_back( MAKE_BINDER(b, e, lightness(), lightness_name) );
    algorithms.push_back( MAKE_BINDER(b, e, dyp(), "dyp") );
    algorithms.push_back( MAKE_BINDER(b, e, nim(), "Nim") );
    algorithms.push_back( MAKE_BINDER(b, e, pts(), "pts") );
    algorithms.push_back( MAKE_BINDER(b, e, epost(), "Eric Postpischil") );
    algorithms.push_back( MAKE_BINDER(b, e, nyar(), "nyarlathotep") );
    algorithms.push_back( MAKE_BINDER(b, e, notbad(), "notbad") );

    {
        std::srand( std::random_device()() );
        std::random_shuffle(begin(algorithms), end(algorithms));
    }

    std::vector< result_t<integer_t> > res;
    for(auto& algo : algorithms)
        res.push_back( algo() );

    auto reference_solution
        = *std::find_if(begin(res), end(res),
                        [](result_t<integer_t> const& p)
                        { return 0 == std::strcmp(lightness_name, p.name); });
    std::cout << "reference solution: "<<reference_solution.name<<"\n\n";

    for(auto const& e : res)
    {
        std::cout << "solution \""<<e.name<<"\":\n";
        auto ms =
            std::chrono::duration_cast<std::chrono::microseconds>(e.duration);
        std::cout << "\tduration: "<<ms.count()/1000<<" ms and "
                                   <<ms.count()%1000<<" microseconds\n";

        std::cout << "\tcomparison to reference solution: ";
        if(e.numbers.size() != reference_solution.numbers.size())
        {
            std::cout << "ouput count mismatch\n";
            break;
        }

        auto mismatch = std::mismatch(begin(e.numbers), end(e.numbers),
                                      begin(reference_solution.numbers)).first;
        if(end(e.numbers) == mismatch)
        {
            std::cout << "exact match\n";
        }else
        {
            std::cout << "mismatch found\n";
        }
    }
}

目前的算法; 注意我用全局函数替换了数字计数器和pow-of-10,所以如果有人优化我们都会受益.

struct lightness
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        using T = typename std::iterator_traits<RaIt>::value_type;

        /**
         * Sorts the array lexicographically.
         *
         * The trick is that we have to compare digits left-to-right
         * (considering typical Latin decimal notation) and that each of
         * two numbers to compare may have a different number of digits.
         *
         * This is very efficient in storage space, but inefficient in
         * execution time; an approach that pre-visits each element and
         * stores a translated representation will at least double your
         * storage requirements (possibly a problem with large inputs)
         * but require only a single translation of each element.
         */
        std::sort(
            b,
            e,
            [](T lhs, T rhs) -> bool {
                // Returns true if lhs < rhs
                // Returns false otherwise
                const auto BASE      = 10;
                const bool LHS_FIRST = true;
                const bool RHS_FIRST = false;
                const bool EQUAL     = false;


                // There's no point in doing anything at all
                // if both inputs are the same; strict-weak
                // ordering requires that we return `false`
                // in this case.
                if (lhs == rhs) {
                    return EQUAL;
                }

                // Compensate for sign
                if (lhs < 0 && rhs < 0) {
                    // When both are negative, sign on its own yields
                    // no clear ordering between the two arguments.
                    //
                    // Remove the sign and continue as for positive
                    // numbers.
                    lhs *= -1;
                    rhs *= -1;
                }
                else if (lhs < 0) {
                    // When the LHS is negative but the RHS is not,
                    // consider the LHS "first" always as we wish to
                    // prioritise the leading '-'.
                    return LHS_FIRST;
                }
                else if (rhs < 0) {
                    // When the RHS is negative but the LHS is not,
                    // consider the RHS "first" always as we wish to
                    // prioritise the leading '-'.
                    return RHS_FIRST;
                }

                // Counting the number of digits in both the LHS and RHS
                // arguments is *almost* trivial.
                const auto lhs_digits = (
                    lhs == 0
                    ? 1
                    : std::ceil(std::log(lhs+1)/std::log(BASE))
                );

                const auto rhs_digits = (
                    rhs == 0
                    ? 1
                    : std::ceil(std::log(rhs+1)/std::log(BASE))
                );

                // Now we loop through the positions, left-to-right,
                // calculating the digit at these positions for each
                // input, and comparing them numerically. The
                // lexicographic nature of the sorting comes from the
                // fact that we are doing this per-digit comparison
                // rather than considering the input value as a whole.
                const auto max_pos = std::max(lhs_digits, rhs_digits);
                for (auto pos = 0; pos < max_pos; pos++) {
                    if (lhs_digits - pos == 0) {
                        // Ran out of digits on the LHS;
                        // prioritise the shorter input
                        return LHS_FIRST;
                    }
                    else if (rhs_digits - pos == 0) {
                        // Ran out of digits on the RHS;
                        // prioritise the shorter input
                        return RHS_FIRST;
                    }
                    else {
                        const auto lhs_x = (lhs / static_cast<decltype(BASE)>(std::pow(BASE, lhs_digits - 1 - pos))) % BASE;
                        const auto rhs_x = (rhs / static_cast<decltype(BASE)>(std::pow(BASE, rhs_digits - 1 - pos))) % BASE;

                        if (lhs_x < rhs_x)
                            return LHS_FIRST;
                        else if (rhs_x < lhs_x)
                            return RHS_FIRST;
                    }
                }

                // If we reached the end and everything still
                // matches up, then something probably went wrong
                // as I'd have expected to catch this in the tests
                // for equality.
                assert("Unknown case encountered");

                // dyp: suppress warning and throw
                throw "up";
            }
        );
    }
};

namespace ndyp
{
    // helper to provide integers with the same number of digits
    template<class T, class U>
    std::pair<T, T> lexicographic_pair_helper(T const p, U const maxDigits)
    {
        auto const digits = count_digits(p);
        // append zeros so that `l` has `maxDigits` digits
        auto const l = static_cast<T>( p  * my_pow10(maxDigits-digits) );
        return {l, p};
    }

    template<class RaIt>
    using pair_vec
        = std::vector<std::pair<typename std::iterator_traits<RaIt>::value_type,
                                typename std::iterator_traits<RaIt>::value_type>>;

    template<class RaIt>
    pair_vec<RaIt> lexicographic_sort(RaIt p_beg, RaIt p_end)
    {
        if(p_beg == p_end) return pair_vec<RaIt>{};

        auto max = *std::max_element(p_beg, p_end);
        auto maxDigits = count_digits(max);

        pair_vec<RaIt> result;
        result.reserve( std::distance(p_beg, p_end) );

        for(auto i = p_beg; i != p_end; ++i)
            result.push_back( lexicographic_pair_helper(*i, maxDigits) );

        using value_type = typename pair_vec<RaIt>::value_type;

        std::sort(begin(result), end(result),
                  [](value_type const& l, value_type const& r)
                  {
                      if(l.first < r.first) return true;
                      if(l.first > r.first) return false;
                      return l.second < r.second; }
                 );

        return result;
    }
}

struct dyp
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        auto pairvec = ndyp::lexicographic_sort(b, e);
        std::transform(begin(pairvec), end(pairvec), b,
                       [](typename decltype(pairvec)::value_type const& e) { return e.second; });
    }
};


namespace nnim
{
    bool comp(int l, int r)
    {
      int lv[10] = {}; // probably possible to get this from numeric_limits
      int rv[10] = {};

      int lc = 10; // ditto
      int rc = 10;
      while (l || r)
      {
        if (l)
        {
          auto t = l / 10;
          lv[--lc] = l - (t * 10);
          l = t;
        }
        if (r)
        {
          auto t = r / 10;
          rv[--rc] = r - (t * 10);
          r = t;
        }
      }
      while (lc < 10 && rc < 10)
      {
        if (lv[lc] == rv[rc])
        {
          lc++;
          rc++;
        }
        else
          return lv[lc] < rv[rc];
      }
      return lc > rc;
    }
}

struct nim
{
    template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, nnim::comp);
    }
};

struct pts
{
        template<class T> static bool lex_less(T a, T b) {
          unsigned la = 1, lb = 1;
          for (T t = a; t > 9; t /= 10) ++la;
          for (T t = b; t > 9; t /= 10) ++lb;
          const bool ll = la < lb;
          while (la > lb) { b *= 10; ++lb; }
          while (lb > la) { a *= 10; ++la; }
          return a == b ? ll : a < b;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, lex_less<typename std::iterator_traits<RaIt>::value_type>);
    }
};

struct epost
{
        static bool compare(int x, int y)
        {
                static const double limit = .5 * (log(INT_MAX) - log(INT_MAX-1));

                double lx = log10(x);
                double ly = log10(y);
                double fx = lx - floor(lx);  // Get the mantissa of lx.
                double fy = ly - floor(ly);  // Get the mantissa of ly.
                return fabs(fx - fy) < limit ? lx < ly : fx < fy;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, compare);
    }
};

struct nyar
{
        static bool lexiSmaller(int i1, int i2)
        {
            int digits1 = count_digits(i1);
            int digits2 = count_digits(i2);

            double val1 = i1/pow(10.0, digits1-1);
            double val2 = i2/pow(10.0, digits2-1);

            while (digits1 > 0 && digits2 > 0 && (int)val1 == (int)val2)
            {
                digits1--;
                digits2--;
                val1 = (val1 - (int)val1)*10;
                val2 = (val2 - (int)val2)*10;
            }
            if (digits1 > 0 && digits2 > 0)
            {
                return (int)val1 < (int)val2;
            }
            return (digits2 > 0);
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, lexiSmaller);
    }
};

struct notbad
{
        static int up_10pow(int n) {
          int ans = 1;
          while (ans < n) ans *= 10;
          return ans;
        }

        static bool compare(int v1, int v2) {
           int ceil1 = up_10pow(v1), ceil2 = up_10pow(v2);
           while ( ceil1 != 0 && ceil2 != 0) {
              if (v1 / ceil1  < v2 / ceil2) return true;
              else if (v1 / ceil1 > v2 / ceil2) return false;
              ceil1 /= 10;
              ceil2 /= 10;
           }
           if (v1 < v2) return true;
           return false;
        }

        template<class RaIt> void operator()(RaIt b, RaIt e)
    {
        std::sort(b, e, compare);
    }
};