shamila08

[code=cplusplus] #include <iostream> #define MAX 100 int num[MAX]; int n; int nrows = 0; using namespace std; void print() { if (num != 0) { for (int j = 0; j <n; j++) cout << num[j]; cout <<"\n"; } } void leftRotate ( int n) {int tmp; for(int k =1; …

Hello..... i got a problem where my output are repeated..... how to solve this problem.... [code=cplusplus] #include <stdio.h> #include <stdlib.h> #include <iostream> #define MAX 100 int num[MAX]; int n; int nrows = 0; int temp =2; using namespace std; void print() { if (num != 0) { for (int i …

hello, dear all. below is my program code. the problem is an half of my output is zero for entries of matrix for n greater than 4. [code=cplusplus] #include <stdio.h> #include <stdlib.h> #include <iostream> #define MAX 100 int num[MAX][MAX]; int n; int nrows = 0; using namespace std; void print(int …

Hello! here c++ program. my problem is i need to find the permutation is even and odd. However the value +1 (even )and -1(odd) is not appear for each permutation. [code=cplusplus]= #include <stdio.h> #include <stdlib.h> #include <iostream> #include <memory.h> #define MAX 100 char num[MAX + 1]; int n; using namespace …

Hello! i got this code. However i'm really not familiar with memcpy function. Can anybody can show me the other way to write that code without using memory function? [code=cplusplus] void rightRotate (int k, int n) { char temp [2*MAX], *saveptr; saveptr = num+k; // cout << "k" << k …

Hello dear all. i need a help where i'm try to calculate execution time for my program. seem i have a problem with how to set it up. [code=cplusplus] #include <iostream> #include <time.h> #include <sys/timeb.h> typedef int tRow[10]; int N, i; int x[100], xi[100]; /* The permutation and its inverse …

hello, dear all [code=cplusplus] #include <iostream> #include <stdio.h> using namespace std; int count = 1; void print( int *arr, int SIZE){ if (arr != 0) { for (int i = 0; i < SIZE; i++) { // i = position cout << arr[i]; } cout <<"\n"; } } void circular_left(int …

Hello, dear all, here is my algorithm. i'm trying to reverse the array but its not function and not loop. [code=cplusplus] #include <iostream> using namespace std; void initial(int *v, int N) { int i; for(i=1; i<=N; i++) v[i] = i; } void write (int *v, int N) { for (int …

Hello, dear all! Below is heap algorithm. the problem is the output is not appear. How to fix it? [code=cplusplus] #include <stdio.h> #include <iostream> using namespace std; int *Value; int N; void read (int N) { int i; for (i = 0; i< N; i++) { cout << Value[i] << …

hello, dear all, i have a slightly problem with my assigment algorithm as follows: [code=cplusplus] void write( int *x, int n) { if (x != 0) { for (int i = 0; i < n; i++) { printf("%4d", x[i] ); } printf("\n"); } } // print void swap(int *x, int …

Hello, dear all. My algorithm as follows. my input is in Code2D.h file. actually i have to put in file 'Code2D.in'. but i dont know have to create it. By the way, my algorithm have no problem with N <= 4. but when i change N = 5, 6 , …

hello, dear all i'm try to find determinant . there is error with ' pow': [code=cplusplus] sum = sum + b[0][p]*pow(-1,p)*determinant(c,m-1); return sum; [/code] after run, the argument was appear as follows: 1>c:\documents and settings\user\my documents\visual studio 2008\projects\deter4\deter4\deter4.cpp(78) : error C2668: 'pow' : ambiguous call to overloaded function 1> c:\program …

Hello, dear all. i have another assignment such as follows: n = 1 print 1 n = 2 print 12 n = 3 print 123 n = 4 print 1234 1243 1423 how to create it by c++?

Hi all.... i have problem with number: example: 1234 is equivalent to 1432 ( based on maths concept), then erase/delete 1432 2143 is equivalent to 2341, then erase/ delete 2341, and so on How to program it in C++

Hello, everybody! i have a problem with permutation array. most algorithm related to permutation, list down all permutation for n!. however, i would like to generate c++ output as follows: 1234 1243 1342 1324 1423 1432 which is we fix the element '1', and cycle others. and soon. i hope …