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- #include "stdafx.h"
- #include <iostream>
- using namespace std;
- int main(){
- //Блок расчета значений функции (1 подпункт задания)
- int a = 1, b = 5, k;
- long double mas[2][100], n;
- cout << "Set n"<< endl;
- cin >> n;
- for (k = 0; (k <= n || k < 100); k++){
- mas[0][k] = a + ((b - a) / n) * k;
- if (mas[0][k] >= 1 && mas[0][k] <= 2)
- mas[1][k] = mas[0][k] - 1;
- else if (mas[0][k] >= 2 && mas[0][k] < 3)
- mas[1][k] = mas[0][k] * mas[0][k] - 3;
- else if (mas[0][k] >= 3 && mas[0][k] <= 5)
- mas[1][k] = 6;
- }
- //Блок вывода в столбики
- int iter = 0;
- int iterEndl = 0;
- while(iter <= k){
- cout << " " << mas[0][iter] << ": " << mas[1][iter] << "\t";
- iterEndl++;
- iter++;
- if(iterEndl == 4){
- cout << endl;
- iterEndl = 0;
- }
- }
- cout << endl << endl;
- //Блок нахождения max, min и другого (2 подпункт задания)
- long double max = mas[1][0], min = mas[1][0];
- for (int i = 0; i <= k; i++){
- if (mas[1][i] < min)
- min = mas[1][i];
- if (mas[1][i] > max)
- max = mas[1][i];
- }
- long double* outMax = &max;
- long double* outMin = &min;
- cout << "min = " << *outMin << endl;
- cout << "max = " << *outMax << endl;
- cout << endl << endl;
- iterEndl = 0;
- for (int i = 0; i <= k; i++){
- if ((((*outMax + *outMin) / 3) >= mas[1][i]) && (mas[1][i] <= *outMax)){
- cout << " " << mas[0][i] << "\t";
- iterEndl++;
- }
- if (iterEndl == 4){
- cout << endl;
- iterEndl = 0;
- }
- }
- return 0;
- }
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