離散數學到c ++

Question

我不確定我應該在這里或在數學網站上詢問，但我會在這里試一試。 下面是作業：用戶輸入10個數字，即域D。 我需要評估語句： For All x, y in D, x<y or y <2x 。 我的問題是什么是x，什么是y？ 如果我輸入1 2 3 4 5 6 7 8 9 10，因為域x = 1，y = 2，依此類推？ 我是否檢查1 <2 || 2 <2（1）？

我編寫了另外三個只使用'x'的語句，但是輸入的十個數字中沒有任何線索，即x和y。 並且它不會要求用戶輸入ax和y只有十個整數。

我會發布我做的其他陳述的代碼。

using std::cout;
using std::cin;

void ASearch(int arr[], int);
void BSearch(int array[], int size);
void CSearch(int array[], int size);

int main(int argc, const char * argv[])
{
    const int SIZE = 10;
    int NumArray[10];
    int num = 0;
    int count = 0;

    while (count < 10)
    {
        cout << "Enter Number "<< count + 1 <<": ";
        cin >> NumArray[count];
        ++count;

    }

    /*if(!isdigit(num))
     {

     for (int i = 0; i < 9; ++i)
     {

     cout << "Enter Number "<< i + 2 <<": ";
     cin >> NumArray[i];

     }
     }else cout << "Error: numbers only";*/

    for(int j = 0; j < SIZE; ++j)
    {
        cout << " " << NumArray[j];
    }

    ASearch(NumArray, SIZE);
    BSearch(NumArray, SIZE);
    CSearch(NumArray, SIZE);


    return 0;
}

void ASearch(int arr[], int size)
{
    int pos = 0;
    int on = -1;
    int oddTrue = 0;
    int oddFalse = 0;

    while (on == -1 && pos < size)
    {
        cout << " \nchecking: " << arr[pos];

        if (arr[pos] % 2 != 0 && arr[pos] > 0)
        {

            cout <<"\nTrue: " <<arr[pos] << " is > 0 and odd";
            ++oddTrue;

        }
        else if(arr[pos] % 2 != 0 && arr[pos] < 0)
        {
            cout <<"\nFalse" << arr[pos] << "is < 0 and odd";
            ++oddFalse;

        }
        else cout << "\nNumber is even";
        ++pos;


    }
    if (oddFalse > 0) cout<< "\n>>>>>>>>>A is FALSE";else cout <<"\n>>>>>>A IS TRUE";
    return;
}

void BSearch(int array[], int size)
{
    int on = -1;
    int pos = 0;
    int count = 0;

    while(on == -1 && pos < size)
    {
        cout << "\n checking "<<array[pos];

        if(array[pos] % 2 != 0  && array[pos] > 10)
        {
            cout << "\nTrue: " << array[pos] <<" is > 10 & odd";
            ++count;

        }else
        {
            cout <<"\nFalse";
            --count;
        }
        ++pos;
    }
    if(count > 1 ) cout << "\n>>>>>>>>>>B IS TRUE"; else cout <<"\n>>>>>>>>>B IS FALSE";
    return;
}


void CSearch(int array[], int size)
{
    int pos = 0;
    int on = -1;
    int TrueCount = 0;
    int FalseCount = 0;

    while (on == -1 && pos < size)
    {
        cout << "\n checking "<<array[pos];

        if(array[pos] % 2 == 0  || array[pos] % 3 == 0)
        {
            cout << "\nTrue: " << array[pos] <<" is divisible by 2 or 3";
            ++TrueCount;

        }
        else if(array[pos] % 2 != 0  || array[pos] % 3 != 0)
        {
            cout <<"\nFalse";
            ++FalseCount;
        }
        ++pos;

    }

    if (TrueCount < 10)
        cout << "\n>>>>C is FALSE"; else cout << "\n>>>>>>>>>>C is TRUE";

}

Answer 1

這意味着x和y都可以采用集合D中的每個值。

在偽代碼中：

for i = 1:10
   for j = 1:10
       x = D[i]
       y = D[j]
       test x < y OR y < 2 * x
   loop
loop