Podstawy Programowania C 02
W S K A Ź N I K I
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zmienna |
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adres x |
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adres y |
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#include <stdio.h>
#include <conio.h>
int main() {
int x,y,pom;
int *wx, *wy;
x=5;
y=7;
printf("\nx=%d\ty=%d\n",x,y);
wx=&x;
wy=&y;
printf("\n\tWydruk przez adres\n");
printf("\nx=%d\ty=%d\n",*wx,*wy);
printf("x?=");
scanf("%d",wx);
printf("y?=");
scanf("%d",wy);
printf("\nx=%d\ty=%d\n",x,y);
printf("\n\tWydruk przez adres\n");
printf("\nx=%d\ty=%d\n",*wx,*wy);
printf("\n\tZamiana przez adres\n");
pom=*wx;
*wx=*wy;
*wy=pom;
printf("\nx=%d\ty=%d\n",x,y);
printf("\n\tWydruk przez adres\n");
printf("\nx=%d\ty=%d\n",*wx,*wy);
fflush(stdin);
getch();
return 0;
}
x=5 y=7
Wydruk przez adres
x=5 y=7
x?=9
y?=2
x=9 y=2
Wydruk przez adres
x=9 y=2
Zamiana przez adres
x=2 y=9
Wydruk przez adres
x=2 y=9
//rysuje linię 10 gwiazdek
#include <stdio.h>
#include <conio.h>
int main() {
int i;
i=1;
while(i<=10) {
printf("*");
i++;
}
fflush(stdin);
getch();
return 0;
}
//rysuje linię x gwiazdek
#include <stdio.h>
#include <conio.h>
int main() {
int i,x;
i=1;
printf("ile gwiazdek ? :");
scanf("%d",&x);
while(i<=x) {
printf("*");
i++;
}
return 0;
}
//rysuje przekątną z x gwiazdek
#include <stdio.h>
#include <conio.h>
int main() {
int i,j,x;
i=1;
printf("ile gwiazdek ? :");
scanf("%d",&x);
while(i<=x) {
j=1;
while(j<i) {
printf(" ");
j++;
}
printf("*\n");
i++;
}
return 0;
}
//rysuje pusty trójkąt prostokątny z x gwiazdek
#include <stdio.h>
#include <conio.h>
//***************************************
int main() {
int i,j,x;
i=1;
printf("ile gwiazdek ? :");
scanf("%d",&x);
printf("*\n");
while(i<=x) {
j=1;
printf("*");
while(j<i) {
printf(" ");
j++;
}
printf("*\n");
i++;
}
i=1;
while(i<=x+2) {
printf("*");
i++;
}
return 0;
}
// rysuje odwrócony pusty trójkąt prostokątny
// z x gwiazdek
#include <stdio.h>
#include <conio.h>
//***************************************
int main() {
int i,j,x;
i=1;
printf("ile gwiazdek ? :");
scanf("%d",&x);
j=x;
while(j>=0) {
printf(" ");
j--;
}
printf("*\n");
while(i<=x) {
j=x;
while(j>=i) {
printf(" ");
j--;
}
j=1;
printf("*");
while(j<i) {
printf(" ");
j++;
}
printf("*\n");
i++;
}
i=1;
while(i<=x+2) {
printf("*");
i++;
}
return 0;
}
//oblicza sumę k liczb
//***************************************
#include <stdio.h>
#include <conio.h>
//***************************************
int main() {
int i,k,suma,pom;
printf("ile liczb ? :");
scanf("%d",&k);
i=1;
suma=0;
while(i<=k) {
printf("podaj liczbe nr %d :",i);
scanf("%d",&pom);
i++;
suma+=pom;
}
printf("\n\nsuma %d liczb wynosi %d",k,suma);
return 0;
}
//oblicza sumę liczb zakończonych znakiem
//***************************************
#include <stdio.h>
#include <conio.h>
//***************************************
int main() {
int i,k,suma,pom;
printf("podaj liczbe nr 1 :",i);
k=scanf("%d",&pom);
i=2;
suma=pom;
while(k!=0) {
printf("podaj liczbe nr %d :",i);
k=scanf("%d",&pom);
i++;
if(k!=0) suma+=pom;
}
i-=2;
printf("\n\nsuma %d liczb wynosi %d",i,suma);
return 0;
}
//oblicza sumę liczb zakończonych znakiem !
//***************************************
#include <stdio.h>
#include <conio.h>
//***************************************
int main(){
int i,k,suma,pom;
char pp;
pp='a';
printf("podaj liczbe nr 1 :",i);
k=scanf("%d",&pom);
i=2;
if(k==0)
scanf("%c",&pp);
suma=pom;
while(pp!='!') {
printf("podaj liczbe nr %d :",i);
k=scanf("%d",&pom);
if(k!=0) {
suma+=pom;
i++;
}
else
scanf("%c",&pp);
}
i--;
printf("\n\nsuma %d liczb wynosi %d",i,suma);
return 0;
}
//znajduje max z liczb zakończonych znakiem !
//***************************************
#include <stdio.h>
#include <conio.h>
int main() {
int i,k,max,pom;
char pp;
pp='a';
printf("podaj liczbe nr 1 :",i);
k=scanf("%d",&pom);
i=2;
if(k==0)
scanf("%c",&pp);
max=pom;
while(pp!='!') {
if(k==0)
printf("\n");
printf("podaj liczbe nr %d :",i);
k=scanf("%d",&pom);
if(k!=0) {
if(max<pom)
max=pom;
i++;
}
else
scanf("%c",&pp);
}
i--;
printf("\n\nmaximum z %d liczb ”,i); printf("wynosi %d",max);
return 0;
}
Zadanie 1.
Algorytm Euklidesa służy do wyznaczania NWD dwóch liczb.
Działa on w następujący sposób:
Mamy 2 liczby a oraz b.
Jeżeli a=b to NWD(a,b)=b,
Jeżeli a≠b to NWD(a,b)=NWD(min(a,b),|a-b|).
Proszę zaimplementować powyższy algorytm.
UWAGA: proszę nie używać rekurencji ale pętli while.
Zadanie 2.
Szybki Algorytm Euklidesa jest modyfikacją Algorytmu Euklidesa.
Działa on w następujący sposób:
Mamy 2 liczby a oraz b.
Jeżeli b=0 to NWD(a,b)=a,
Jeżeli b≠0 to NWD(a,b)=NWD(b, (a mod b))
Proszę zaimplementować powyższy algorytm.
UWAGA: proszę nie używać rekurencji ale pętli while.
Zadanie 3.
Algorytm rosyjskich wieśniaków służy do obliczania iloczynu dwóch liczb.
Działa on w następujący sposób:
Mamy 2 liczby a oraz b.
Tworzymy dwie kolumny. Następnie w kolejnych wierszach wpisujemy :
w pierwszym wierszu :
w kolumnie a wartość a,
w kolumnie b wartość b,
w następnych wierszach:
w kolumnie a część całkowitą uzyskaną z podzielenia liczby z poprzedniego wiersza przez 2,
w kolumnie b liczbę uzyskaną z pomnożenia liczby z poprzedniego wiersza przez 2.
Postępujemy tak dopóki liczba w kolumnie a jest większa od 1.
Następnie sumujemy liczby z kolumny b, dla których ich odpowiedniki w kolumnie a są nieparzyste.
a=11 |
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b=6 |
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66 |
//NWD - algorytm Euklidesa
#include <stdio.h>
#include <conio.h>
int main(){
int x, y, p1, p2,p3;
printf("podaj a :");
scanf("%d",&x);
printf("podaj b :");
scanf("%d",&y);
p1=x;
p2=y;
while(p2!=p1) {
if(p1>p2)
p1=p1-p2;
else
p2=p2-p1;
}
printf("\n\nNWD(%d,%d)=%d",x,y,p1);
fflush(stdin);
getch();
return 0;
}
//NWD - szybki algorytm Euklidesa
#include <stdio.h>
#include <conio.h>
int main(){
int x, y, p1, p2,p3;
printf("podaj a :");
scanf("%d",&x);
printf("podaj b :");
scanf("%d",&y);
p1=x;
p2=y;
while(p2!=0) {
p3=p2;
p2=p1%p2;
p1=p3;
}
printf("\n\nNWD(%d,%d)=%d",x,y,p1);
fflush(stdin);
getch();
return 0;
}
//mnożenie - algorytm rosyjskich wieśniaków
#include <stdio.h>
#include <conio.h>
int main(){
int x, y, p1, p2,p3;
printf("podaj a :");
scanf("%d",&x);
printf("podaj b :");
scanf("%d",&y);
p1=x;
p2=y;
p3=0;
while(p1!=1) {
if(p1%2)
p3+=p2;
p1=p1/2;
p2*=2;
}
p3+=p2;
printf("\n\n%d*%d=%d",x,y,p3);
fflush(stdin);
getch();
return 0;
}
//mnożenie - algorytm rosyjskich wieśniaków
//wersja wielokrotna
#include <stdio.h>
#include <conio.h>
int main(){
int x, y, p1, p2,p3,stop;
stop=1;
while(stop!=0) {
printf("\n\npodaj a :");
stop=scanf("%d",&x);
if(stop!=0) {
printf("podaj b :");
scanf("%d",&y);
p1=x;
p2=y;
p3=0;
while(p1!=1) {
if(p1%2)
p3+=p2;
p1=p1/2;
p2*=2;
}
p3+=p2;
printf("\n\n%d*%d=%d",x,y,p3);
fflush(stdin);
getch();
system("cls");
}
else
getchar();
}
system("cls");
printf("\n\nK O N I E C");
fflush(stdin);
getch();
return 0;
}
Zadanie 1. Proszę napisać program, który
1.wczyta z klawiatury liczbę linii
2.wydrukuje obrazek:
*
**
***
****
*****
******
Zadanie 2. Proszę napisać program, który
1.wczyta z klawiatury:
a)liczbę elementów w 1 wierszu
b)liczbę wierszy
c)różnicę pomiędzy ilością elementów pomiędzy kolejnymi wierszami
2.policzy sumę elementów tak zdefiniowanej struktury danych, wczytywanych z klawiatury
Przykład 1:
Liczba elementów w 1 wierszu:
3
liczba wierszy:
5
różnica:
2
Wygląd danych
5, 7, 3
8, 6, 4, 5, 9
13,21, 2, 5, 6, 8, 7
Suma=109
Przykład 2:
Liczba elementów w 1 wierszu:
7
liczba wierszy:
5
różnica:
-2
Wygląd danych
13,21, 2, 5, 6, 8, 7
8, 6, 4, 5, 9
5, 7, 3
Suma=109
Zadanie 3.
Rekurencyjna definicja ciągu Fibonacciego ma postać:
a0=1
a1=1
an=an-1+an-2
Proszę napisać program obliczjący iteracyjnie n-ty wyraz ciągu Fibonacciego.
Zadanie 4. Proszę napisać program, który
wczyta z klawiatury podzielniki a1 i a2 oraz liczby x1 i x2, a następnie wczyta dane z klawiatury. Wczytywanie kończy się gdy liczba danych podzielnych przez a1 jest równa x1 lub liczba danych podzielna przez a2 jest równa x2. Program ma wyliczyć i wydrukować różnicę sumy wszystkich danych pomniejszoną o iloczyn danych podzielnych przez a1 i podwojoną sumę danych podzielnych przez a2.
Zadanie 5. Proszę napisać program, który
1.wczyta z klawiatury liczbę linii
2.liczbę gałęzi
3.wydrukuje obrazek dla 3 linii i 4 gałęzi:
*
***
*****
*
***
*****
*
***
*****
*
***
*****
Zadanie 6*.
Rekurencyjna definicja ciągu ma postać:
a0=1
a1=1
a2=2
an=2an-1+3an-2 +an-3- n dla n>2
Proszę napisać program obliczjący iteracyjnie n-ty wyraz tego ciągu.
Zadanie 7*.
Proszę napisać program, który wykonuje operację szybkiego potęgowania na dwóch liczbach całkowitych (modyfikacja algorytmu rosyjskich wieśniaków ).
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