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Thread: Çàäà÷à íà Ñ++
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26th March 2010 20:11 #1Registered User
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Çàäà÷à íà Ñ++
Çäðàâåéòå íà âñè÷êè. Òúé êàòî íå ðàçáèðàì ìíîãî îò ïðîãðàìèðàíå ñå äîïèòâàì äî âàñ. Íÿêîé ùå ìîæå ëè äà ìè ïîìîãíå ñ åäíà çàäà÷à ? Ùå ñå ðàäâàì íà âñÿêàêúâ âèä ïîìîù îò âàøà ñòðàíà
Åòî è çàäà÷àòà:
Çàä. Ïðåäñòàâåòå ôóíêöèèòå àáñîëþòíà ñòîéíîñò
abs, labs, fabs ñ àðãóìåíòè îò òèï ñúîòâåòíî
int, long, double, ôóíêöèèòå ceil, floor–
çà ïðåîáðàçóâàíå íà ðåàëíî ÷èñëî â öÿëî ÷èñëî,
ôóíêöèÿòà sqrt – êâàäðàòåí êîðåí è ôóíêöèÿòà pow–
ñòåïåíóâàíå. Çà âñÿêà ãðóïà îò ôóíêöèè äà áúäå
ïîñî÷åí è ñúîòâåòíèÿ çàãëàâåí ôàéë ( <cstdlib>
èëè <cmath>), êîéòî äà áúäå âêëþ÷åí â ïðîãðàìàòà.
#include<iostream.h>
#include<cstdlib>
#include<cmath>
using namespace std;
int main ()
{
int m=5, n=-7;
long k=5, l=-7;
double x=5.2, y=-7.7;
double a=-5.2, b=7.7;
cout<<"Absolute value functions"<<endl;
cout<<"|m|="<<abs(m)<< endl //cstdlib
<<"|n|="<<abs(n)<< endl;
cout<<"|k|="<<labs(k)<< endl //cstdlib
<<"|l|="<<labs(l)<< endl;
cout<<"|x|="<<fabs(x)<< endl //cmath
<<"|y|="<<fabs(y)<< endl<<endl;
cout<<"ceil function - round up <cmath>"<< endl;
cout<<"ceil(5.2) ="<<ceil(5.2)<< endl;
cout<<"ceil(-7.7)="<<ceil(-7.7)<< endl;
cout<<"ceil(-5.2)="<<ceil(-5.2)<< endl;
cout<<"ceil(7.7)="<<ceil(7.7)<< endl<<endl;
cout<<"floor function - round down <cmath>"<< endl;
cout<<"floor(5.2) ="<<floor(5.2)<< endl;
cout<<"floor(-7.7)="<<floor(-7.7)<< endl;
cout<<"floor(-5.2)="<<floor(-5.2)<< endl;
cout<<"floor(7.7)="<<floor(7.7)<< endl<<endl;
cout<<"sqrt function <cmath>" << endl;
cout<<"sqrt (16)= "<<sqrt(16) <<endl;
cout<<"sqrt (16.0)="<<sqrt (16.0) <<endl<<endl;
cout<<"pow function <cmath>" endl;
cout<<"pow(5.2,3)="<<pow(5.2,3)<< endl;
cout<<"pow(-5.2,3.0)="<<pow(-5.2,3.0)<< endl;
cout<<"pow(5.2,-3)="<<pow(5.2,-3)<< endl;
cout<<"pow(-5.2,-3.0)="<<pow(-5.2,-3.0)<< endl;
double z=pow(sqrt(4),2) + sqrt(pow(2,4))<< endl;
cout<<"z="<<z<< endl;
return 0;
}
Çàäà÷à 6. Äà ñå íàïèøå èçðàç íà åçèêà C++, êîéòî äà èçðàçÿâà:
à) ïåðèìåòúða íà êâàäðàò ñ ëèöå, ðàâíî íà a;
á) ëèöåòî íà ðàâíîñòðàíåí òðèúãúëíèê ñ ïåðèìåòúð, ðàâåí íà p.
Çàä. Äà ñå íàïèøå ïðîãðàìà, êîÿòî äà ñúäúðæà Ñ++ ôóíêöèÿ, êîÿòî äà ïðåñìÿòà ëèöåòî íà òðèúãúëíèê ïî äàäåíè ñòðàíè a,b,c.(ôóíêöèè, êîèòî âðúùàò ñòîéíîñò)
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26th March 2010 21:50 #2Registered User
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Îòíîñíî çàäà÷à 6. Ïðåäëàãàì ïúðâî äà ñè íàïèøåø ìàòåìàòè÷åñêèòå èçðàçè çà äâåòå ïîäóñëîâèÿ. Îò òàì íà áàçàòà íà èçïèñàíîòî â ïúðâàòà çàäà÷à íå áè òðÿáâàëî äà èìàø ïðîáëåìè ñúñ ñúñòàâÿíåòî íà èçðàçèòå íà C++.
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26th March 2010 22:35 #3Registered User
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26th March 2010 23:08 #4
Å çà à ïîä óñëîâèå íå å ëè -> p = (sqrt(a))*4 ; ?
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26th March 2010 23:15 #5
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26th March 2010 23:33 #6
Òö. Çà âòîðîòî ôîðìóëàòà å : S=(P/6)*sqrt(((p/3)^2)-((p/6)^2))
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27th March 2010 00:33 #7
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27th March 2010 12:23 #8
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27th March 2010 09:06 #9Registered User
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Çíà÷è çà ïîäòî÷êà à) äà íàïèøà p = (sqrt(a))*4, à çà ïîäòî÷êà á) s=(sqr(p))/(12*sqrt(3)) èëè S=(P/6)*sqrt(((p/3)^2)-((p/6)^2)) ?
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27th March 2010 11:08 #10Registered User
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Êàòî íà÷àëî ìîæåø äà îïèòàø äà îñúçíàåø êàêâè ñà ìàòåìàòè÷åñêèòå ôóíêöèè â C++ è êàêâà ðàáîòà âúðøàò. Ñëåä òîâà ùå ñòèãíåø äî îòãîâîð íà âúïðîñèòå ñè ñàìîñòîÿòåëíî.
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27th March 2010 12:31 #11Registered User
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Áúðêàø - â íåãîâàòà ôîðìóëà sqr îçíà÷àâà ïîâäèãàíå íà êâàäðàò, à íå êîðåí êâàäðàòåí, êîåòî å çàïèñàë êàòî sqrt. Íå ïîìíÿ íà êîé åçèê áåøå òàêà, ìîæå áè íà "Ïàñêàë". Ôîðìóëàòà ìó å ïî÷òè âÿðíà, òúé êàòî âèñî÷èíàòà â ðàâíîñòðàíåí òðèúãúëíèê å ðàâíà íà ñòðàíàòà ïî (êîðåí îò 3)/2, à ñòðàíàòà å ÿñíî êîëêî îò ïåðèìåòúðà

Ïî÷òè âÿðíà, çàùîòî êàòî ñå âãëåäàõ, âèäÿõ, ÷å å îáúðêàë ìÿñòîòî íà âúïðîñíèÿ (êîðåí îò 3) - òðÿáâà äà å â ÷èñëèòåëÿ, íå â çíàìåíàòåëÿ.
Ðåäàêöèÿ äåí ïî-êúñíî - íàèñòèíà ôîðìóëàòà å âÿðíà. Àç ñúì ñå îáúðêàë, ïîíåæå íå ñúì îáúðíàë âíèìàíèå, ÷å â çíàìåíàòåëÿ ñúùî èìà òðîéêè, êîèòî "ñå ñúêðàùàâàò" ñ (êîðåí îò 3) â ÷èñëèòåëÿ, ïðè êîåòî òîé îòèâà â çíàìåíàòåëÿ.Last edited by bsb; 28th March 2010 at 11:15.
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27th March 2010 13:15 #12
Òîâà íå ñúì ãî äîãëåäàë. Ïðè ñìÿòàíå ñ Ð íà êâàäðàò ñå ïîëó÷àâà ñúùîòî ëèöå êàòî ïî äðóãàòà ôîðìóëà. Ïëþñ òîâà êîðåíà ñè å áàø êúäåòî òðÿáâà. Ïðîáâàé äà ïîëó÷èø ôîðìóëàòà íà ëèñò õàðòèÿ è ùå ñå óáåäèø
sqrt(((P/6)^3)*(P/2)) ïàê å ñúùîòî
Last edited by haibusa2005; 27th March 2010 at 13:34.
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27th March 2010 22:05 #13Registered User
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Íÿêîé ìîæå ëè äà íàïèøå òîâà, êîåòî å íåîáõîäèìî êúì ïîäòî÷êà à) è á)
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27th March 2010 22:08 #14Êîéòî ìè öúêíå ÷åðâåí ïàëåö, äà ìó îìåêíå ÷óðêàòà!




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