Laboratoriya mashg’uloti-6

Laboratoriya mashg’uloti-6

Mavzu: Muhandislik masalalarini yechishda dasturlash tizimlari yordamida grafika va multimedia imkoniyatlaridan foydalanish

Masalaning qo’yilishi:

Quyida berilgan tarmoqlanuvchi algoritmlarning dasturini tuzing.




_{𝑌 = {} ^√|𝑛₁ ∙ 𝑛₂^| 𝑎𝑔𝑎𝑟 𝑛₁𝑛₂ < 0.2

^√|𝑛₁ + 𝑛₂^| 𝑎𝑔𝑎𝑟 𝑛₁𝑛₂ ≥ 0.1

Bunda 𝑛₁ = −10, 𝑛₂ = 3.

𝑌𝑒𝑐ℎ𝑖𝑚𝑖𝑛𝑖 10⁻³ 𝑎𝑛𝑖𝑞𝑙𝑖𝑘𝑑𝑎 𝑐ℎ𝑖𝑞𝑎𝑟𝑖𝑛𝑔. 𝑁𝑎𝑡𝑖𝑗𝑎: 𝑌 = 5.477
Ishdan maqsad.

Tarmoqlanuvchi algoritmlar va ular ustida amalllar bajarish ko’nikmalarini oshirish. Dasturini tuzish. if shart, case tanlash operatorlarini tog’ri qo’llashni o’rganish, tarmoqlanuvchi algoritmga doir masalalarni dasturini tuzishni o’rganish.

Nazariy qism.

If va if else operatorlari, ?: amali.

Dastur bajarilishining birorta qadamida qandaydir shartni tekshirish natijasiga ko’ra boshqaruvni dasturning u yoki bu bo’lagiga uzatish mumkin(Tarmoqlanuvchi algoritm). Tarmoqlanishni amalga oshirish uchun shartli operatordan foydalaniladi. Agar algoritm qadamlari ketma-ket bajarilish jarayonida qandaydir shartga bo’gliq ravishda o’zgarsa, bunday algoritm tarmoqlanuvchi algoritm deb nomlanadi. Shart bu mantiqiy ifoda bo’lib, faqat rost yoki yolg’on qiymatni qabul qiladi. Agar shart rost bo’lsa Xa, yolg’on bo’lsa Yo’q tarmog’i bo’yicha algoritm qadami davom

etadi.

If operatori. If qandaydir shartni rostlikka tekshirish natijasiga ko’ra dasturda tarmoqlanishni amalga oshiradi:

If (< shart >) ;
Bu yerda har qanday ifoda bo’lishi mumkin, odatda u taqqoslash amali bo’ladi.

Agar shart 0 qiymatidan farqli yoki rost (true) bo’lsa, bajariladi, aks holda, ya’ni shart 0 yoki yolg’on (false) bo’lsa, hech qanday amal bajarilmaydi va boshqaruv if operatoridan keyingi operatorga o’tadi (agar u mavjud bo’lsa).

C++ tilining qurilmalari operatorlarni blok ko’rinishida tashkil qilishga imkon beradi. Blok - ‘{‘ va ‘}’ belgi oralig’iga olingan operatorlar ketma-ketligi bo’lib, u kompilyator tomonidan yaxlit bir operator deb qabul qilinadi. Blok ichida e’lon operatorlari ham bo’lishi mumkin va ularda e’lon qilingan o’zgaruvchilar faqat shu blok ichida ko’rinadi (amal qiladi), blokdan tashqarida ko’rinmaydi. Blokdan keyin ‘;’ belgisi qo’yilmasligi mumkin, lekin blok ichidagi har bir ifoda ‘;’ belgisi bilan yakunlanishi shart.

Shart operatorining if - else ko’rinishi quyidagicha:

if () ; else ;

Bu yerda 0 qiymatidan farqli yoki true bo’lsa, , aks holda bajariladi. Misol tariqasida diskreminantni hisoblash usuli yordamida 𝑎𝑥 ² + 𝑏𝑥 + 𝑐 = 0 ko’rinishidagi kvadrat tenglama ildizlarini topish masalasini ko’raylik:

Variantlar. Topshiriqlar. 1-vazifa.

Quyidagi variantlarda berilgan tarmoqlanuvchi algoritmlar dasturini tuzing.

	Variantlar		Kirish	Chiqish
1	𝑚^2 𝑛 + 1 − 𝑐, 𝑎𝑔𝑎𝑟 𝑛 + 1 > 0 𝑌 = { 2 2 (𝑚 + 𝑛) + 𝑐𝑚 , 𝑎𝑔𝑎𝑟 𝑛 + 1 ≤ 0		m=1 n=2 c=3	Y=0.000
			m=2 n=3 c=7	Y=6.000
2	^1 , 𝑎𝑔𝑎𝑟 𝑎 < 5 𝑌 = { ^√9+𝑥^2𝑏𝑢 𝑦𝑒𝑟𝑑𝑎 𝑎 = 𝑑^2 + 𝑏 ∙ 𝑠𝑖𝑛𝛼 , 𝑎𝑔𝑎𝑟 𝑎 ≥ 5 𝑐𝑑 ; 𝑐^2−𝑑^2		b=1 c=2 d=3 x=4	Y=0.999
			b=2 c=3 d=7 x=-6	Y=-1.952
3	7𝑥 ^2 − 3𝑎𝑏 − 5𝑎𝑏, 𝑎𝑔𝑎𝑟 𝑎 ≥ 0 𝑍 = { ; 15𝑎 − 7𝑏, 𝑎𝑔𝑎𝑟 𝑎 < 0		a=1 b=2	Z=47.000

4			x=3
			a=-2 b=9 x=7	Z=-93.000
5	𝑌 = ^𝑎2^+𝑏2 + √𝑎^2 + 𝑥, 𝑎𝑔𝑎𝑟 𝑥 ≥ 0 2 2 { ^𝑐 𝑏𝑢 𝑦𝑒𝑟𝑑𝑎 𝑥 = ^𝑎^−𝑏; ^𝑠𝑖𝑛𝑥^+𝑏, 𝑎𝑔𝑎𝑟 𝑥 < 0 𝑐^2−𝑏 𝑎−𝑏		a=1 b=2 c=3	Y=-1.090
			a=2 b=3 c=7	Y=-2.170
6	(𝑛𝑚^2 + 𝑑)^2, 𝑎𝑔𝑎𝑟 𝑑 > 1 𝑌 = { 𝑑 ; 2 2 , 𝑎𝑔𝑎𝑟 𝑑 ≤ 1 𝑛 +𝑚		m=1 n=2 d=3	Y=25.000
			m=4 n=7 d=0.2	Y=0.003
7	^𝑎𝑥2 , 𝑎𝑔𝑎𝑟 𝑎 < 9 𝑍 = { ^𝑏−^1; (𝑎 + 1)^3 + 𝑐𝑥 ^3, 𝑎𝑔𝑎𝑟 𝑎 ≥ 9		a=1 b=2 c=3 x=4	Z=16.000
			a=12 b=3 c=7 x=9	Z=7300.00 0
8	^𝑎3 , 𝑎𝑔𝑎𝑟 𝑎 > 0 𝑋 = { 3+𝑎 𝑏𝑢𝑛𝑑𝑎 𝑎 = ^𝑏2 ^−𝑐2; 2 √|^𝑎^+^2| , 𝑎𝑔𝑎𝑟 𝑎 ≤ 0 1+𝑎		b=4 c=-2 d=8	X=0.750
			b=4 c=-3 d=7	X=0.250
9	𝑥 ^√| | − √𝑥, 𝑎𝑔𝑎𝑟 𝑥 > 3 𝑍 = 𝑥 + 1 (^𝑙𝑛^|^𝑥^|)^3, 𝑎𝑔𝑎𝑟 𝑥 ≤ 3 { 𝑥		x=2	Z=0.042
			x=3.7	Z=1.066
10	𝑥𝑒^𝑥𝑠𝑖𝑛𝑥 , 𝑎𝑔𝑎𝑟 𝑥 ≥ 0 ^𝐾^=^{^1𝑙𝑛^3^|𝑥^|, 𝑎𝑔𝑎𝑟 𝑥 < 0^; 3		x=3	K=8.503
			x=-7	K=2.456
11	ln(𝑥 + √𝑥 ^2 + 9) , 𝑎𝑔𝑎𝑟 𝑥 ≥ 0 𝑌 = { ; 𝑡𝑔^3𝑥, 𝑎𝑔𝑎𝑟 𝑥 < 0		x=3	Y=1.980
			x=-7.2	Y=0.308
12	^𝑎 , 𝑎𝑔𝑎𝑟 𝑎 > 5 𝑌 = { ^𝑎2 ^+𝑥; 1 2𝑎3+𝑠𝑖𝑛𝑎 𝑎𝑔𝑎𝑟 𝑎 ≤ 5		a=4 x=9	Y=0.008
			a=7 x=1	Y=0.140

13	√𝑘, 𝑎𝑔𝑎𝑟 𝑠𝑖𝑛𝑘 ≤ 0.2 𝑌 = { 1 , 𝑎𝑔𝑎𝑟 𝑠𝑖𝑛𝑘 > 0.2 √^𝑘		k=7	Y=0.378
			k=1.4	Y=0.845
14	𝑠𝑖𝑛^2 ^(2𝑥^)− 𝑐𝑜 𝑠^2𝑥, 𝑎𝑔𝑎𝑟 𝑥 > 0 ^𝑌^=^{1 ^; 2 , 𝑎𝑔𝑎𝑟 𝑥 ≤ 0.3 ^𝑥^−√^𝑥		x=0.5	Y=- 0.062
			x=0.7	Y=0.386
15	^1 ^(3𝑥 ^2 − 1^), 𝑎𝑔𝑎𝑟 𝑥 > 0.4 𝑃 = { ^2 ; ^1 ^(5𝑥 ^3 − 3𝑥^), 𝑎𝑔𝑎𝑟 𝑥 ≤ 0.4 2		x=7.3	P=7 9.435
			x=0.12	P=-0.176
16	𝑥 ^2 + 4, 𝑎𝑔𝑎𝑟 𝑥 < 10 𝑎^2−𝑏 𝑌 = { 3 𝑏𝑢𝑛𝑑𝑎 𝑥 = 𝑐 ; 𝑥 − 7, 𝑎𝑔𝑎𝑟 𝑥 ≥ 10		a=1 b=2 c=3	Y=4.11
			a=-2 b=3 c=7	Y=4.02
17	√15𝑎^2 + 21𝑏^2, 𝑎𝑔𝑎𝑟 𝑎 > 𝑏 𝑌 = { ; √15𝑏^2 + 21𝑎^2 , 𝑎𝑔𝑎𝑟 𝑎 ≤ 𝑏		a=7 b=3	Y=30.397
			a=1.2 b=3.8	Y=15.711
18	𝑙𝑛^|2𝑥 − 3𝑧^2^|, 𝑎𝑔𝑎𝑟 ^|𝑥^|< 5𝑧 𝑌 = { | 2 | | | ; 𝑙𝑛 2𝑥 − 3𝑧 , 𝑎𝑔𝑎𝑟 𝑥 > 5𝑧		x=-10 z=3	Y=3.850
			x=6.3 z=-7	Y=4.609
19	sin^(5𝑘 + 3𝑚^), 𝑎𝑔𝑎𝑟 ^|𝑘^|> ^|𝑚^| ^𝑃^=^{cos^(5𝑘 + 3𝑚^), 𝑎𝑔𝑎𝑟 ^|𝑘^|≤ ^|𝑚^|^;		k=5 m=-6	Y=0.754
			k=-4.2 m=3.7	Y=0.458
20	√^2𝑘1^−^7𝑘2,^𝑎𝑔𝑎𝑟^𝑘1𝑘2^<^1 𝑌 = { ; √^2𝑘1^+^7𝑘2,^𝑎𝑔𝑎𝑟^𝑘1𝑘2^≥^1		𝑘1 = 0.02 𝑘2 = 4	Y=5.288
			𝑘1 = 9 𝑘2 = 7	Y=8.185
21	^4𝑟+3𝑚 , 𝑎𝑔𝑎𝑟 ^|𝑟^|> ^|𝑚^|+ ^1 2 2 𝑌 = { 𝑟 +𝑚 2 ; ^|𝑟 − 𝑚^|, 𝑎𝑔𝑎𝑟 ^|𝑟^|≤ ^|𝑚^|+ ^1 2		m=3 r=5	Y=0.853
			m=6.5 r=7	Y=0.500
22	^𝑎 , 𝑎𝑔𝑎𝑟 𝑎 > 5 𝑌 = { ^𝑎2 ^+𝑥; 1 3 𝑎𝑔𝑎𝑟 𝑎 ≤ 5 2𝑎 +𝑠𝑖𝑛𝑎		a=4 x=9	Y=0.008
			a=7 x=1	Y=0.140
23	√𝑘, 𝑎𝑔𝑎𝑟 𝑠𝑖𝑛𝑘 ≤ 0.2 𝑌 = { 1 , 𝑎𝑔𝑎𝑟 𝑠𝑖𝑛𝑘 > 0.2 √^𝑘		k=7	Y=0.378
			k=1.4	Y=0.845

24	𝑠𝑖𝑛^2 ^(2𝑥^)− 𝑐𝑜 𝑠^2𝑥, 𝑎𝑔𝑎𝑟 𝑥 > 0 ^𝑌^=^{1 ^; 𝑥2−√𝑥 , 𝑎𝑔𝑎𝑟 𝑥 ≤ 0.3		x=0.5	Y=-0.062
			x=0.7	Y=0.386
25	^1 ^(3𝑥 ^2 − 1^), 𝑎𝑔𝑎𝑟 𝑥 > 0.4 𝑃 = { ^2 ; ^1 ^(5𝑥 ^3 − 3𝑥^), 𝑎𝑔𝑎𝑟 𝑥 ≤ 0.4 2		x=7.3	P=79.435
			x=0.12	P=-0.176
26	𝑥 ^2 + 4, 𝑎𝑔𝑎𝑟 𝑥 < 10 𝑎^2−𝑏 𝑌 = { 3 𝑏𝑢𝑛𝑑𝑎 𝑥 = ; 𝑥 − 7, 𝑎𝑔𝑎𝑟 𝑥 ≥ 10 ^𝑐		a=1 b=2 c=3	Y=4.11
			a=-2 b=3 c=7	Y=4.02
27	√15𝑎^2 + 21𝑏^2, 𝑎𝑔𝑎𝑟 𝑎 > 𝑏 𝑌 = { ; √15𝑏^2 + 21𝑎^2 , 𝑎𝑔𝑎𝑟 𝑎 ≤ 𝑏		a=7 b=3	Y=30.397
			a=1.2 b=3.8	Y=15.711
28	𝑙𝑛^|2𝑥 − 3𝑧^2^|, 𝑎𝑔𝑎𝑟 ^|𝑥^|< 5𝑧 𝑌 = { | 2 | | | ; 𝑙𝑛 2𝑥 − 3𝑧 , 𝑎𝑔𝑎𝑟 𝑥 > 5𝑧		x=-10 z=3	Y=3.850
			x=6.3 z=-7	Y=4.609
29	sin^(5𝑘 + 3𝑚^), 𝑎𝑔𝑎𝑟 ^|𝑘^|> ^|𝑚^| ^𝑃^=^{cos^(5𝑘 + 3𝑚^), 𝑎𝑔𝑎𝑟 ^|𝑘^|≤ ^|𝑚^|^;		k=5 m=-6	Y=0.754
			k=-4.2 m=3.7	Y=0.458