Question

编辑：我目前正在开展一个项目。我几乎完成了，虽然我似乎无法弄清楚如何使用非重复随机数填充我的向量（矩阵）。我得到随机数字，但其中一些正在重复。我尝试实现一个方法来检查推送到向量的数字是否已经存在，但这不起作用。

有没有简单的方法可以解决这个问题？

#include <iostream>
#include <iomanip>
#include <vector>
#include <cstdlib>
#include <ctime>

using namespace std;

/*
The purpose of this program is to create a matrix using 2D vectors, then decide
if it is a perfect matrix or not.
*/

void matrix(int );

int main()
{
    int input;
    char again = 'y';

    cout << "Welcome to my perfect matrix program. The function of the program is"
            "\nto:"
            "\n"
            "\n   1. Allow the user to enter the size of the perfect matrix, such as"
            "\n      N. N>=2."
            "\n   2. Create a 2 D vector array of size N x N."
            "\n   3. Populate the 2 D vector array with distinct random numbers."
            "\n   4. Display the sum for each row, column, and diagonals then"
            "\n      determine whether the numbers in N x N vector array are perfect"
            "\n      matrix numbers.";
    cout << endl << endl;

    while(again == 'y' || again == 'Y')
    {
        cout << "\nEnter the size of the matrix : ";
        cin >> input;

        while(input < 2)
        {
            cout << "\nError *** Incorrect input - You entered a number < 2"
                    "\nEnter a Positive integer Number >= 2: ";
            cin >> input;
        }

        matrix(input);

        cout << "\nWould you like to find another perfect matrix?"
                "\nEnter y || Y for yes or n || N for no: ";
        cin >> again;

        if(again == 'N' || again == 'n')
        {
            cout << "\nThis perfect matrix algorithm is implemented by B"
                    "\nFebruary 13th - 2017\n";
        }

        while(again != 'Y' && again != 'y' && again != 'N' && again != 'n')
        {
            cout << "\nError *** Invalid choice - Must enter y|Y or n|N" << endl;
            cout << "\nWould you like to find another perfect matrix?"
                    "\nEnter y || Y for yes or n || N for no: ";
            cin >> again;
        }
    }

    return 0;
}

void matrix(int value)
{
    int N = value;
    bool isPerf = true;
    int sum = 0,
        i,
        j;

    int diagSum1 = 0,
        diagSum2 = 0;

    vector<vector<int> >arr;

    srand(time(0));

    for(i = 0; i < N; i++)
    {
        vector<int> temp;
        bool check;
        int digit;
        for(j = 0; j < N; j++)
        {
            do
            {
                digit = rand()%8;
                check = true;
                for(int k = 0; k < j; k++)
                {
                    if(digit == temp.front())
                    {
                        check = false;
                        break;
                    }
                }
            }while(!check);
            temp.push_back(digit);
        }
        arr.push_back(temp);
    }

    cout << endl;
    cout << "The perfect matrix that is created for size " << value << ":";
    cout << endl << endl;

    for(i = 0; i < arr.size(); i++)
    {
        for(j = 0; j < arr[i].size(); j++)
        {
            cout << arr[i][j];
            cout << "   ";
        }
        cout << endl << endl;
    }

    for(i = 0; i < arr.size(); i++)
    {
        for(j = 0; j < arr[i].size(); j++)
        {
            sum += arr[i][j];
        }
    }
    int perf = sum/3;

    cout << "The perfect number is: " << perf << endl << endl;

    for(i = 0; i < arr.size(); i++)
    {
        int rowSum = 0;
        for(j = 0; j < arr[i].size(); j++)
        {
            rowSum += arr[i][j];
        }
        cout << "Sum of numbers in Row # " << i + 1 << " = " << rowSum;
        if(rowSum != perf)
        {
            isPerf = false;
        }
        cout << endl;
    }
    cout << endl;


    for(i = 0; i < arr.size(); i++)
    {
        int colSum = 0;
        for(j = 0; j < arr[i].size(); j++)
        {
            colSum += arr[j][i];
        }

        cout << "Sum of numbers in Column # " << i + 1 << " = " << colSum;
        if(colSum != perf)
        {
            isPerf = false;
        }
        cout << endl;
    }
    cout << endl;


for(i = 0; i < arr.size(); i++)
{
        diagSum1 += arr[i][i];
        diagSum2 += arr[i][arr.size() - i - 1];
}

cout << "Sum of numbers in first diagonal    =   " << setw(3) << diagSum1 << endl;
if(diagSum1 != perf)
    {
        isPerf = false;
    }
cout << "Sum of numbers in second diagonal   =   " <<  setw(3) << diagSum2 << endl;
if(diagSum2 != perf)
    {
        isPerf = false;
    }

cout << endl;

if(isPerf)
{
    cout << "The above is a perfect matrix" << endl;
}
else
    cout << "The above is not a perfect matrix" << endl;

cout << endl;

}

Answer 1

一个选项是创建一个临时集，供您填充随机数。插入元素时检查集合的大小，当大小等于所需的不同元素的数量时，迭代集合以检索值并填充矩阵。

...
std::set<int> distinct;
while (distinct.size() < input) {
        int r_num = /* random number generation */
        distinct.insert(r_num);
}
std::set<int>::iterator it;
for (it = distinct.begin(); it != distinct.end(); it++) {
        /* method of adding to matrix */ = *it;
}
...

非重复随机数2D矢量（C ++）

1 个答案: