Kam unë ca detyra... në C++

[The Pathfinder]

Kam une ca detyra...!

Per programim:

1 -
A- Te formohet vektori G(m) duke shumezuar katroret e anetareve perkates te vektorit A(m) me vlera absolute te anetareve me indeks te njejte te vektorit B(m). Njekohesisht anetaret e G(m) qe jane me te medhenj se nje numer i dhene x, duhet te pjesetohen me indekset perkatese.

B- Te numerohet numri k i anetareve te vektorit G(m) me vlera absolute midis 3 dhe 15, duke shtypur njekohesisht anetaret me vlera jashte ketij diapazoni si dhe indekset perkatese.

C- Gjej anetaret me vlera pozitive dhe negative ne kete vektor.

D- Rradhit sipas madhesise vlerat numerike te anetareve te ketij vektori duke filluar nga me e madhja.

2-
A- Te formohet matrica A[m][n] kur dihen m dhe n duke pjesetuar elementet perkates me numrin pozitiv x.

B- Te ndertohet nje algoritem qe gjen shumen e elementeve te matrices.

C- Te kalohen ne vektor sipas shtyllave elementet e matrices.

D- Te formohet prej saj nje matricve e re duke hequr kolonen e dyte.

[hot_prinz]

Detyra 1:

#include "stdafx.h"
#include<iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    //deklarimi i matricave statike a dhe b
    int a[]={1,-2,3,4,-5,6,7,8,-9};
    int b[]={9,-8,7,-6,5,-4,3,-2,1};
    //deklarimi dhe dimensionimi i matrices g 
    //sa elemente gjinden ne matricen a
    int g[sizeof(a)/sizeof(a[0])];
    int k=0;
    cout<<"\n Detyrat 1\n\n A.)\n\n";
    for(int i = 0; i<(sizeof(a)/sizeof(a[0]));i++)
    {
        //llogaritja e katroreve te elementeve
        //te matrices a shumezuar me vlerat
        //absolute te elementeve matrices b
        g[i] = (a[i] * a[i]) * b[i];
        //nese marrim se x = 50 atehere
        if(g[i]>50)
        {
            //pjestojme elementet e matrices g 
            //me indeksin e elementeve
            g[i] = g[i]/i;
        }
        cout<<" matrica g["<< i <<"] = "<< g[i]<<"\n";
    }
    cout<<"\n B.)\n\n";
    for(int i = 0; i<(sizeof(g)/sizeof(g[0]));i++)
    {
        //numroji elementet e matrices g  
        //nese gjinden mes vlerave 3 dhe 15
        if(g[i]>3 && g[i]<15)
        {
            k++;
        }
        else
        {
            //shfaq elementet tjere jashte vlerave
            cout<<" matrica g["<<i<<"] = "<<g[i]<<"\n";
        }
    }
    cout<<" "<<k<<" elemente te matrices g[] gjinden mes 3 dhe 15!\n";
    cout<<"\n C.)\n\n";
    for(int i = 0; i<(sizeof(g)/sizeof(g[0]));i++)
    {
        //gjejm vlerat negative dhe pozitive te matrices g
        if(g[i]>0)
            cout<<" matrica g["<<i<<"] ka vlere pozitive = ("<<g[i]<<")\n";
        if(g[i]<0)
            cout<<" matrica g["<<i<<"] ka vlere negative = ("<<g[i]<<")\n";
    }
    cout<<"\n D.)\n\n";
    //radhitja e elementeve te matrices g
    //sipas algoritmit radhites [Bubble Sort]
    int i, j, flag = 1;
    int temp;
    int arrayLength = (sizeof(g)/sizeof(g[0]));
    for(i = 1; (i <= arrayLength) && flag; i++)
    {
        flag = 0;
        for (j=0; j < (arrayLength -1); j++)
        {
            if (g[j+1] > g[j])
            { 
                temp = g[j];
                g[j] = g[j+1];
                g[j+1] = temp;
                flag = 1;
            }
        }
    }
    for(int i = 0; i<(sizeof(g)/sizeof(g[0]));i++)
    {
        cout<<" matrica g["<<i<<"] = "<<g[i]<<"\n";
    }
    cout<<" matrica g[] pas radhitjes me algoritmin [Bubble Sort]:\n";
    cout<<"\n fund...\n\n\n";
    system("pause");
    return 0;
} 

[hot_prinz]

Detyra 2:

#include "stdafx.h"
#include<iostream>
#include "stdlib.h"
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    const int M = 5;
    const int N = 4;

    int A[M][N];
    int x = 4;
    int m=0;
    int n=0;
    int shuma=0;

    for(m=0;m<M;m++)
    {
        for(n=0;n<N;n++)
        {
            //mbushja e matrices A[][] me vlera 
            //randome prej 50 deri ne 100
            A[m][n] = 50 + ( rand() % 100 ) ;
            A[m][n] = A[m][n] / x;
        }
    }
    cout << "\n\n Detyra 2 \n\n\n A)\n\n";
    cout << "       m1  m2  m3  m4  m5\n\n";
    for (n=0; n<N; n++)
    {
        cout <<" n"<<n<< "  ";
        for (m=0; m<M; m++)
        {
            cout << "  " << A[m][n];
        }
        cout << endl;
    }
    cout << "\n\n B)\n\n\n";
    for (m=0; m<M; m++)
    {
        for (n=0; n<N; n++)
        {
            shuma += A[n][m];
        }
    }
    cout << " shuma e te gjithe elementeve te matrices eshte = " << shuma;
    cout << "\n\n\n C)\n\n\n";
    int a[M*N];
    for(m=0;m<M;m++)
    {
        for(n=0;n<N;n++)
        {
            a[n] = A[m][n];
            cout << " vektori a[" << n << "] = " << a[n] << "\n";
        }
    } 
    cout << "\n\n\n D)\n\n";
    cout << "       m1  m3  m4  m5\n\n";
    for (n=0; n<N; n++)
    {
        cout <<" n"<<n<< "  ";
        for (m=0; m<M; m++)
        {
            if(m!=1)
            {
                cout << "  " << A[m][n];
            }
        }
        cout << endl;
    }
    cout << "\n\n\n fund...!\n\n\n";
    system("pause");
    return 0;
} 

shpresoj se i kam kuptuar mire detyrat, sepse nuk ishin aq te qarta!

[The Pathfinder]

Shume faleminderit.
Je i nderuar!

[The Pathfinder]

Po ne C, si do zgjidheshin keto ushtrime?

Ku ndryshon C nga C++ ne kode, dhe scripte?!

Ju faleminderit per ndihmen tuaj!

[hot_prinz]

ne keto dy detyra, duhet te pershtatet vetem funksioni cout (C++) ne printf(C). Testuar me DEV C++:

Detyra 1 ne C:

#include <stdio.h>
#include <stdlib.h> 

int main(int argc, char *argv[])
{
    
  int a[]={1,-2,3,4,-5,6,7,8,-9};
  int b[]={9,-8,7,-6,5,-4,3,-2,1};
  int g[sizeof(a)/sizeof(a[0])]; 
  int k=0; 
  int i = 0;
    printf("\n Detyra 1 \n\n A.)\n\n"); 
    for(i = 0; i<(sizeof(a)/sizeof(a[0]));i++) 
    { 
        g[i] = (a[i] * a[i]) * b[i]; 
        if(g[i]>50) 
        { 
            g[i] = g[i]/i; 
        } 
        printf(" matrica g[%1d] = %4d\n",i,g[i]); 
    } 
    
    printf("\n B.)\n\n"); 
    i=0;
    for(i = 0; i<(sizeof(g)/sizeof(g[0]));i++) 
    { 
        if(g[i]>3 && g[i]<15) 
        { 
            k++; 
        } 
        else 
        { 
            printf(" matrica g[%1d] = %4d\n",i,g[i]);
        } 
    } 
    printf(" %1d elemente te matrices g[] gjinden mes 3 dhe 15!\n",k);
    i=0;
    printf("\n\n C.)\n\n"); 
    for(i = 0; i<(sizeof(g)/sizeof(g[0]));i++) 
    { 
        if(g[i]>0) 
            printf(" matrica g[%1d] ka vlere pozitive = (%4d)\n",i,g[i]);
        if(g[i]<0)  
            printf(" matrica g[%1d] ka vlere negative = (%4d)\n",i,g[i]);
    } 
    
    printf("\n D.)\n\n"); 
    i=0;
    int j, flag = 1; 
    int temp; 
    int arrayLength = (sizeof(g)/sizeof(g[0])); 
    for(i = 1; (i <= arrayLength) && flag; i++) 
    { 
        flag = 0; 
        for (j=0; j < (arrayLength -1); j++) 
        { 
            if (g[j+1] > g[j]) 
            {  
                temp = g[j]; 
                g[j] = g[j+1]; 
                g[j+1] = temp; 
                flag = 1; 
            } 
        } 
    } 
    i=0;
    for(i = 0; i<(sizeof(g)/sizeof(g[0]));i++) 
    { 
        printf(" matrica g[%1d] = (%4d)\n",i,g[i]);
    } 
    printf(" matrica g[] pas radhitjes me algoritmin [Bubble Sort]:\n"); 
    printf("\n fund...\n\n\n"); 

  system("PAUSE");    
  return 0;
} 

dhe Detyra 2 ne C:

#include <stdio.h>
#include <stdlib.h> 

int main(int argc, char *argv[])
{
    const int M = 5; 
    const int N = 4; 

    int A[M][N]; 
    int x = 4; 
    int m=0; 
    int n=0; 
    int shuma=0; 

    for(m=0;m<M;m++) 
    { 
        for(n=0;n<N;n++) 
        { 
            A[m][n] = 50 + ( rand() % 100 ) ; 
            A[m][n] = A[m][n] / x; 
        } 
    } 
    
    printf( "\n\n Detyra 2 \n\n\n A)\n\n"); 
    printf( "       m1  m2  m3  m4  m5\n\n"); 
    for (n=0; n<N; n++) 
    { 
        printf(" n%1d  ",n); 
        for (m=0; m<M; m++) 
        { 
            printf( "  %2d", A[m][n]); 
        } 
        printf("\n"); 
    } 
    
    printf("\n\n B)\n\n\n"); 
    for (m=0; m<M; m++) 
    { 
        for (n=0; n<N; n++) 
        { 
            shuma += A[n][m]; 
        } 
    } 
    printf(" shuma e te gjithe elementeve te matrices eshte = %4d", shuma); 


    printf("\n\n\n C)\n\n\n"); 
    int a[M*N]; 
    for(m=0;m<M;m++) 
    { 
        for(n=0;n<N;n++) 
        { 
            a[n] = A[m][n]; 
            printf(" vektori a[%2d] = %4d\n",n, a[n]); 
        } 
    }  

    printf("\n\n\n D)\n\n"); 
    printf("       m1  m3  m4  m5\n\n"); 
    for (n=0; n<N; n++) 
    { 
        printf(" n%2d  ",n); 
        for (m=0; m<M; m++) 
        { 
            if(m!=1) 
            { 
                printf("  %2d", A[m][n]); 
            } 
        } 
        printf("\n"); 
    } 
    printf("\n\n\n fund...!\n\n\n"); 

  system("PAUSE");    
  return 0;
}