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四位数随机数,无数字重复

[英]Four digit random number without digit repetition

Is there any way you can have a 4 digit number without repetition - eg not 1130 but 1234 ? 有什么办法可以不重复地使用4位数字-例如不是1130而是1234 I read std::random_shuffle could do this but it would only swap the numbers in between. 我读了std::random_shuffle可以做到这一点,但它只会在两者之间交换数字。

#include <iostream>
#include <string>
#include <cstdlib>
#include <ctime>
#include <random>

unsigned seed = static_cast<size_t>(std::chrono::system_clock::now().time_since_epoch().count());

using namespace std;

class Player {
private:
    string playername;

public:
    void setName(string b) {
        cout << "Please enter your name:" << endl;
        getline(cin, b);
        playername = b; 
    }

    string getName () {
        return playername;
    }
};

class PasswordGuessingGame {
private:
    std::mt19937 random_engine;
    std::uniform_int_distribution<size_t> random_generator;

public:
    PasswordGuessingGame():
        random_engine(seed),
        random_generator(1000,9999)
    { 
    }

    int getNumber () {
        return random_generator(random_engine);
    }
};

int main () {
    Player newgame;
    PasswordGuessingGame b;
    newgame.setName("");

    cout << newgame.getName() << " " <<  "password " << b.getNumber() <<  endl;
    }

One possibility is to generate a string containing the digits, and to use the C++14 function std::experimental::sample() 一种可能是生成包含数字的字符串,并使用C ++ 14函数std::experimental::sample()

#include <iostream>
#include <random>
#include <string>
#include <iterator>
#include <experimental/algorithm>

int main() {
std::string in = "0123456789", out;
do {
    out="";
    std::experimental::sample(in.begin(), in.end(), std::back_inserter(out), 4, std::mt19937{std::random_device{}()});
    std::shuffle(out.begin(), out.end(), std::mt19937{std::random_device{}()});
  } while (out[0]=='0');
  std::cout << "random four-digit number with unique digits:"  << out << '\n';
}

Edit: 编辑:

Changed to prevent a result that starts with a 0. Hat tip to @Bathsheba who indicated that this could be a problem. 进行了更改,以防止结果以0开头。@ Bathsheba的帽子提示指出这可能是个问题。

I think you need to generate each digit separately. 我认为您需要分别生成每个数字。 For example you have array from 0 to 9 with 0..9 digits. 例如,您有一个从0到9的数组,其中包含0..9位数。 For first digit you generate number from 0 to 9 and pick up digit from this array. 对于第一个数字,您会生成0到9之间的数字,并从此数组中提取数字。 Then you swap this array element t with last element of array. 然后将此数组元素t与数组的最后一个元素交换。 For second digit you generate number form 0 to 8. And so on. 对于第二个数字,您生成的数字格式为0到8。依此类推。

I don't see a problem with std::random_shuffle : 我看不到std::random_shuffle有问题:

#include <iostream>
#include <string>
#include <algorithm>
#include <random>
#include <chrono>

int main()
{
    std::string s;    
    std::generate_n(std::back_inserter(s), 10,
        []() { static char c = '0'; return c++; });
    // s is now "0123456789"

    std::mt19937 gen(std::random_device{}());

    // if 0 can't be the first digit
    std::uniform_int_distribution<size_t> dist(1, 9);
    std::swap(s[0], s[dist(gen)]);

    // shuffle the remaining range
    std::shuffle(s.begin() + 1, s.end(), gen); // non-deprecated version

    // convert only first four
    auto x = std::stoul(s.substr(0, 4));
    std::cout << x << std::endl;
}

Live on Coliru 住在科利鲁

If you can count how many valid (ie acceptable) sequences exist, and you can devise a bijective function that maps from this counter to each valid sequence instance then things become trivial. 如果您可以计算存在多少有效(即可接受)序列,并且可以设计出一个双射函数,将这个计数器映射到每个有效序列实例,那么事情就变得微不足道了。

If I understand your example correctly, you want to generate a random 4 digit sequence ABCD (representing an integer in the range [0,9999]) where digits A , B , C and D are different from one another. 如果我正确理解了您的示例,则希望生成一个随机的4位数序列ABCD(表示[0,9999]范围内的整数),其中ABCD的数字彼此不同。

There are 5040 such valid sequences: 10 * 9 * 8 * 7. 有5040个这样的有效序列:10 * 9 * 8 * 7。

Given any integer in the range [0, 5039], the following function will return a valid sequence (ie one in which each digit is unique), represented as an integer: 给定[0,5039]范围内的任何整数,以下函数将返回一个有效的序列(即每个数字都是唯一的序列),以整数表示:

int counter2sequence(int u) {
  int m = u/504;
  u %= 504;
  int h = u/56;
  u %= 56;
  int t = u/7;
  u %= 7;

  const int ih = h;
  const int it = t;

  if (ih >= m) ++h;
  if (it >= ih) ++t;
  if (t >= m) ++t;
  if (u >= it) ++u;
  if (u >= ih) ++u;
  if (u >= m) ++u;

  return ((m*10 + h)*10 + t)*10 + u;
}

Eg 例如

counter2sequence(0) => 0123
counter2sequence(5039) => 9876

Another method, using only standard data and numeric algorithms. 另一种方法,仅使用标准数据和数字算法。

#include <random>
#include <array>
#include <iostream>
#include <numeric>

template<class Engine>
int unrepeated_digits(int ndigits, Engine &eng) {
    std::array<int, 10> digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
    auto add_digit = [](auto x, auto digit) {
        return x * 10 + digit;
    };
    std::shuffle(std::begin(digits), std::end(digits), eng);
    return std::accumulate(std::begin(digits), std::next(std::begin(digits), ndigits), 0, add_digit);
}

int main() {
    std::random_device rnd;
    std::default_random_engine eng(rnd());

    for (int i = 0; i < 10; ++i)
        std::cout << unrepeated_digits(4, eng) << std::endl;
}

example output: 示例输出:

7623
3860
9563
9150
3219
8652
4789
2457
1826
9745

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