Guruh talabasi Shodmonov Anvarning Algoritimlarni loyihalash fanidan 2-mustaqil ishi

Dastur kodi :

#include

#include

using namespace std;

void matrix_print(int a[10][10], int m, int n)

{

// matritsani jadval shaklida chiqarish

for (int i = 0; i < m; i++)

{

for (int j = 0; j < n; j++)

{

cout << a[i][j] << "\t";

}

cout << "\n";

}

}

int satr_max(int a[], int n)

{

// massivning eng katta elementini aniqlash

int max = a[0];

for (int i = 1; i < n; i++)

if (max < a[i]) max = a[i];

return max;

}

int main()

{

int m, n, a[10][10];

cout << "Satrlar sonini kiriting \nm="; cin >> m;

cout << "Ustunlar sonini kiriting \nn="; cin >> n;

cout <<"Massiv elementlarini kiriting \n";

for (int i = 0; i < m; i++)

for (int j = 0; j < n; j++)

cin >> a[i][j];

cout << "Kiritilgan matritsa\n";

//funksiyaga matritsa, satrlar va ustunlar soni jo'natiladi

matrix_print(a, m, n);

for (int i = 0; i < m; i++)

{

// funksiyaga i-satrning 0-elementi adresini

// va elementlar sonini jo'natamiz

cout << i << "-satrning eng kattasi=" <<

satr_max(&a[i][0], n);

cout << endl;

}

return 0;

}



2-masala.

Quyidagi funksiyani to’rtburchaklar, Trapetsiya va Simpson formulalari yordamida taqribiy hisoblash dasturini tuzing

Bizning misolda bo’lgani uchun,to’g’ri o’rtburchaklar formulasiga asosan, quyidagi natijani hosil qilamiz.

X=0.1; =0.9901

X=0.2; =0.9615

X=0.3; =0.9174

X=0.4; =0.8621

X=1.0 =0.5000

7.5998

Trapetsiya usulida kodi:

#include

#include

using namespace std;

double function(double x)

{

return (sin(x)/(1+x*x));

}

int main()

{

double a,b,S=0, s1;

int n;

cout<<"integral chegarasini kiriting"<

cin>>a>>b;

cout<<"Qadamlar sonini kiriting:"<

cin>>n;

s1=a+0.1;

while (s1

{

S+=function(s1);

s1+=0.1;

}

S=(a+b)/2+S;

S=S*fabs(b-a)/n;

cout << S;

return 0; }

Turtburchak usuli:

#include

using namespace std;

double function(double x)

{

return (sin(x)/(1+x*x));

}

int main()

{

double a,b,S=0, sum1,n;

cout<<"qadamlar sonini kiriting:"<>n;

cout<<"integral chegarasini kiriting"<

cin>>a>>b;

sum1=a+0.1;

while (sum1

{

S+=function(sum1);

sum1+=0.1;

}

S=S*fabs(b-a)/n;

cout << S;

return 0;

}