Kod: 001 Fan: iaktt yn 2024 yil




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Axborot xati konf. ADU, Mayers- Briggs qo\'shimcha, 36-qo\'shma qaror, 27.04.2022, Oila tushunchasi, uning turlari va shakillari, fHy1I56Pj1m1Sqci4f9q3e28B9S0AiBM, dars ishlanma, 11-21-ALGORITMIK TILLAR VA DASTURLASH, Мустақил ишни ташкиллаштириш, Иқтибослик учун, Документ Microsoft Word, Calendar plan-RAQAMLI VA AXBOROT TEXNOLOGIYALARI (2), статья, Исмаилова Н С , Шагазатов У У Жахон иқтисодиёти ва халқаро (1), A5

Kod: 001 Fan: IAKTT YN «___»_________2024 yil
F.I.Sh. ___________________________________ Guruh №__________ Imzo:__________

“Kompyuter ilmlari va dasturlashtirish” kafedrasi mudiri O.A.Kayumov __________________ «___» ____ 2023 y.

I. Savol: Raqamli iqtisodiyotda axborot - kommunikatsiya texnologiyalarining o‘rni va vazifalari haqida ma’lumot bering? 7 ball

II. Iqtisodiy axborotni tahlil qilish uchun ma’lumotlarni vizualizatsiya qilish vositalari haqida ma’lumot bering? 7 ball

III. Biznesdagi avtomatlashtirilgan axborot tizimlarining roli haqida ma’lumot bering? 7 ball

IV. Sanoq sistemalari ustida amallar bajaring? 7 ball
1) 10001012X10 2) 12410X2 3) FB16X10 4) 3548X10 5) DF16 +B716 =X16 6) 5678+4568=X8

V. Dasturlashda quyidagi berilganlarni ishlang? 22 ball

1) arifmetik amallarni hisoblang? 3 ball
int a=9, b=11, c=15;
a+=a++ - ++b*++c/b%a; b+=b++/++c*++a%b;
c+=++b%a*c+ ++c; a=_____; b=_____; c=______;
2) mantiqiy amallarni hisoblang? 4 ball
int a=7, b=9, c=13; bool d;
a+=((a|b)+(c&b))+b|c; b+=((a|c)+(c&b))+b^c;
c+=((a&c)+(c&b))+b^a; d=(aa=______; b=____; c=___; d=_______;
3) razryadli amallarni hisoblang? 3 ball
int a=14, b=13, c=15;
a+=((a|b)+(c&b))+b^c; b+=((a&c)+(c^b))+b|c;
c+=((a&c)+(c&b))+b^a; a>>=2; b>>=2; c<<=3;
a=_____; b=_____; c=_____;
4) Shartga nisbatan a va c ni aniqlang? 4 ball
int a=9, b=3, c=7, d=(a+b)%c;

if(!(a++>++b)|(c++<++d)){a+=((a^b)<(c&d))?b:d; }
else{c+=((d|b)<(c&b))?a:c;} a-?______ c-? _____
5) Takrorlanish jarayonini tahlil qiling? 4 ball
int a=9, b=4, c=13;
if (a > c){ while (a > c){ a -= 3; b += (a-c); c += 2; } }
else { for (int i = a; i < c; i++){
b += (a % 3 + c % 5); } } b?______
6) Berilganlar: x=3.74x10-2, y=-0.825, z=0.16x102,
dasturini tuzing? 4 ball
Yakuniy nazorat variantlari 2023-yil 4-dekabrdagi kafedra yig’ilishida muhokama qilingan va 8-sonli yig’ilish bayoni bilan tasdiqlangan.

1
#include


using namespace std;


int main()


{
int a=9, b=11, c=15;
a+=a++ - ++b*++c/b%a;
b+=b++/++c*++a%b;
c+=++b%a*c+ ++c;

cout << "a=" << a << endl;


cout << "b=" << b << endl;
cout << "c=" << c << endl;
return 0;
}

2
#include


using namespace std;


int main()


{
int a=7, b=9, c=13;
bool d;
a+=((a|b)+(c&b))+b|c;
b+=((a|c)+(c&b))+b^c;
c+=((a&c)+(c&b))+b^a;
d=(a

cout << "a=" << a << endl;


cout << "b=" << b << endl;
cout << "c=" << c << endl;
cout << "d=" << d << endl;
return 0;
}
3
#include

using namespace std;


int main()


{
int a=14, b=13, c=15;
a+=((a|b)+(c&b))+b^c;
b+=((a&c)+(c^b))+b|c;
c+=((a&c)+(c&b))+b^a;
a>>=2; b>>=2; c<<=3;

cout << "a=" << a << endl;


cout << "b=" << b << endl;
cout << "c=" << c << endl;

return 0;


}
4
#include

using namespace std;


int main()


{
int a=9, b=3, c=7, d=(a+b)%c;
if(!(a++>++b)|(c++<++d)){
a+=((a^b)<(c&d))?b:d;
}else{
c+=((d|b)<(c&b))?a:c;
}
cout << "a=" << a << endl;
cout << "c=" << c << endl;
return 0;
}
5
#include

using namespace std;


int main()


{
int a=9, b=4, c=13;
if (a > c){
while (a > c){
a -= 3; b += (a-c); c += 2;
}
}else {
for (int i = a; i < c; i++){
b += (a % 3 + c % 5);
}
}
cout << "b=" << b << endl;

return 0;


}
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