The basics of programming in c++ for beginners

Tasks: arithmetic operations in C ++

If you have already read the articleArithmetic operations in C ++ you can begin to practice.

1. Common task:  Given the four-digit number (for example 5678), display the numbers in reverse order of which is the number of member. That is, we should see on the screen 8765. Tip: to take from among the individual numbers, should be applied to the modulo 10.

2. The site of almost any commercial bank, you can find the so-called Deposit calculator, which allows people to, not wishing to go into the formula for calculating interest rates, to know how much they will receive. To do this, they just fill in certain fields, press the button and see the result. This is a simple program, which has already been able to write each one of you. So, a task: The user enters the amount of the deposit and the number of months of keeping money in the bank. It is necessary to calculate and show the screen profit from the deposit in a month,  for the entire term of the deposit, and the total amount payable at the end of the period.  Currency let it be – U.S. dollar. Interest rate – 5% APR.  The formula for calculating percent per month–                      SumDeposit * (interest rate / 100)  / daysperyear * dayspermonths.

Perhaps you have any questions about the solution of tasks – ask them in the comments!

151 thoughts on “Tasks: arithmetic operations in C ++

  1. #include
    using namespace std;

    int main(){
    unsigned long int a,c=0;
    cin>>a;
    while (a!=0){
    c*=10;
    c=c+a%10;
    a/=10;
    }
    cout<<c;
    return 0;
    }

  2. #include
    using namespace std;

    int main()
    {
    setlocale(LC_CTYPE, “Russian”);
    float depozit = 0, srok = 0;
    cout <> depozit;
    cout <> srok;
    cout << "прибыль в месяц:" << depozit * 0.05 / 365 * 30 << "$" << endl;
    cout << "прибыль за весь срок:" << depozit * 0.05 / 365 * 30 * srok << "$" << endl;
    cout << "итоговая выплата:" << depozit * 0.05 / 365 * 30 * srok + depozit << "$" << endl;
    return 0;
    }

  3. Пойду против правил. Зачем писать большой код, если можно так:

    #include
    using namespace std;

    int main()
    {
    char a, b, c, d;
    cin >> a >> b >> c >> d;
    cout << d << c << b << a;
    }

    1. Ахахахааххаахах гениально.
      Немного доработал ваш код.

      #include
      using namespace std;

      int main ()
      {
      int a=5;
      int b=6;
      int c=7;
      int d=8;
      cout << "Данно целое число: " < <a << b << c << d << endl;
      cout << "Число наизнанку: " << d << c << b << a << endl;
      }

  4. #include
    using namespace std;

    void magazin() { //zadacha iz purecodecpp.com/archives/433
    int m = 0, sh = 0, co = 0;
    int sum;
    int mRub = 79, shRub = 124, coRub = 735;
    int sumRub;
    int menunum;
    system(“color 2″);
    for (menunum = 0; menunum < 4;) {
    cout << "\t\t~ DOBRO POJALOVAT' V MAGAZIN CITY 17 ~\n\n";
    cout <> DOBAVITV KORZINU: \n====================================================================\n”;
    cout <> 1. MOLOKO\t\t\t79 rub.\t\t—-\t” << m << "\tsht.\n";
    cout <> 2. SHOKOLADKA\t\t124 rub.\t—-\t” << sh << "\tsht.\n";
    cout <> 3. DEVYATOK YAITS\t\t735 rub.\t—-\t” << co << "\tsht.\n";
    cout <> 4. KUPITS DOSTAVKOY\t3500 rub.\t—-\t-\t—-\n>>\n>> “;
    cin >> menunum;
    switch (menunum) {
    case 1:
    m++;
    break;
    case 2:
    sh++;
    break;
    case 3:
    co++;
    break;
    case 4:
    sum = m + sh + co;
    mRub *= m;
    shRub *= sh;
    coRub *= co;
    sumRub = mRub + shRub + coRub + 3500;
    cout << "\n====================================================================\n";
    cout <> ITOGO:\t\t\t” << sumRub << "* rub.\t—-\t" << sum <> *s uchetom dostavki :)\n\n”;
    system(“pause”);
    default:
    menunum = 0;
    }
    if (menunum != 4) {
    menunum = 0;
    }
    system(“cls”);
    }
    }

    void bank() { //zadacha iz purecodecpp.com/archives/433
    int depo = 0, mes = 0, prib = 0;
    int procmes = 0, procgod = 0;
    int menunum;
    system(“color 2″);
    for (menunum = 0; menunum < 3;) {
    cout << "\t\t~ DOBRO POJALOVAT' V BANK CITY 17 ~\n\n";
    cout <> DEPOZITNIY KALKULATOR: \n============SPESHITE! PROTSENTNAYA STAVKA 0.2% GODOVIH!!!========\n>>\n”;
    cout <> 1. SUMMA DEPOZITA\t\t\t” << depo << "\trub.\n";
    cout <> 2. SKOL’KO MESYATSEV HRANIT’\t\t” << my << "\tmesyatsev\n";
    cout <> 3. POSCHITATPRIBIL’…\n>>\n>> “;
    cin >> menunum;
    switch (menunum) {
    case 1:
    system(“cls”);
    cout << "\t\t~ DOBRO POJALOVAT' V BANK CITY 17 ~\n\n";
    cout <> DEPOZITNIY KALKULATOR: \n============SPESHITE! PROTSENTNAYA STAVKA 0.2% GODOVIH!!!========\n>>\n”;
    cout <> depo;
    break;
    case 2:
    system(“cls”);
    cout << "\t\t~ DOBRO POJALOVAT' V BANK CITY 17 ~\n\n";
    cout <> DEPOZITNIY KALKULATOR: \n============SPESHITE! PROTSENTNAYA STAVKA 0.2% GODOVIH!!!========\n>>\n”;
    cout <> my;
    break;
    case 3:
    procmes = depo * (0.2 / 100) / 365 * 30;
    procgod = procmes * 12;
    cout << "\n====================================================================\n";
    cout <> PRIBILSOSTAVIT:\n>>\n”;
    cout <> V MESYATS:\t\t\t” << procmes << "\trub.\n";
    cout <> V GOD:\t\t\t” << procgod << "\trub.\n\n";
    system("PAUSE");
    default:
    menunum = 0;
    }
    if (menunum != 3) {
    menunum = 0;
    }
    system("cls");
    }
    }

    int main()
    {
    int menuitem = 0;
    system("color 2");
    cout << "\t\t~ DOBRO POJALOVAT' V 'My HeLLo WORLD!' ~\n\n";
    cout <> VIBRATPUNKT: \n====================================================================\n”;
    cout <> 1. MAGAZIN\n”;
    cout <> 2. BANK\n”;
    cout <> 99. VIHOD\n>>\n>> “;
    cin >> menuitem;
    switch (menuitem) {
    case 1:
    system(“cls”);
    magazin();
    break;
    case 2:
    system(“cls”);
    bank();
    break;
    case 99:
    system(“exit”);
    default:
    menuitem = 0;
    }
    return 0;
    }

  5. #include
    using namespace std;
    int main()
    {
    int num;
    cin >> on one;
    for (on one; on one != 0; num /= 10)
    cout << on one % 10;
    return 0;

    }

  6. по формуле расчета процентов консультировался с экономистом, тут программа считает сложный процент.

    #include
    #include
    using namespace std;
    int main() {
    double b, d, x, t, c;
    int a;

    cout << "enter summ depozit " <> x;
    cout << "enter kol-vo mounse " <> a;
    cout << "enter % stavku " <> c;

    c /= 100;
    t = 1 + c / 12;
    b = pow(t, a) ;
    b *= x;
    d = (b – x) / a;

    cout << "summa vklada itog : " << b << endl;
    cout << "zarabotok v mes : " << d << endl;

    return 0;
    }

  7. short a=0;

    cin >> a;

    cout << a % 10;
    cout << (a / 10) % 10;
    cout << (a / 100) % 10;
    cout << (a / 1000) % 10;

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