def count_odds(values): 695 >>> count_odds([[111, 165, 207]]) [3] >>> count_odds([[1, 2], [8], [5, 6, 7]]) [1,0, 2] 695
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- #define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #define NAME_LEN 25 #define MAX_PARTS 100 struct part { int number; char name[NAME_LEN + 1]; int on_hand; double price; } inventory[MAX_PARTS]; int num_parts = 0; int find_part(int number); void insert(void); void search(void); void update(void); void print(void); int read_line(char str[], int n); int main(void) { char code; for (;;) { printf("Enter operation code: "); scanf(" %c", &code); while (getchar() != '\n') /* skips to end of line */ ; switch (code) { case 'i': insert(); break; case 's': search(); break; case 'u': update(); break; case 'p': print(); break; case 'q': return 0; default: printf("Illegal code\n"); } printf("\n"); } } int find_part(int number) { int i; for (i = 0; i < num_parts; i++) if…Match the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eax* .gcd(91, 287) =gcd(91, 14) yes No
- Dec2Hex function : def decimal_to_hex(number): hex_map = {0: '0', 1: '1', 2: '2', 3: '3', 4: '4', 5: '5', 6: '6', 7: '7', 8: '8', 9: '9', 10: 'A', 11: 'B', 12: 'C', 13: 'D', 14: 'E', 15: 'F'} hex_digits = [] while number > 0: hex_digits.append(hex_map[number % 16]) number //= 16 # Reverse the order of the hex digits and join them to form the final hex string hex_string = ''.join(reversed(hex_digits)) return f'0x{hex_string}' if hex_string else '0x0' Perform a Profile of your Dec2Hex function. Write a function that takes a timedate object as a parameter andcalculates the number of years from NOW to the time in the timedateobject. Add unit-testing code to your Dec2Hex exercise, and then perform aUnit test of the updated Dec2Hex code.in c language !! typedef struct {long restaurant_id;char restaurant_name[10];char description[120];double rate;char cuisine[30]; opening_year;long capacity;char city[30];char address[60];char owner[30];} RECORD_t, *RECORD; RECORD delete (unsigned long restaurant_id, RECORD *restaurant_array, unsigned long *p_size){int rec;int index;// index <- find the index of the RECORD with given restaurant_id in restaurant_array ????? (I DIDNT)if (index is valid)rec=restaurant_array[index];elserec=NULL;return rec;}bit_flag.c #include <stdio.h> #include <stdlib.h> #include "bit_flags.h" typedef struct bit_flags { int size; int capacity; int *bit; } Bit_flags; BIT_FLAGS bit_flags_init_number_of_bits(int number_of_bits) { Bit_flags* pBit_flags; if(number_of_bits > 0) // The given number is positive { pBit_flags = (Bit_flags*)malloc(sizeof(Bit_flags)); if(pBit_flags != NULL) { pBit_flags->capacity = sizeof(int); pBit_flags->size = number_of_bits; pBit_flags->bit = (int*)malloc(sizeof(int)); if(pBit_flags->bit != NULL) { *pBit_flags->bit = 0; } else { printf("Could not set memory for bit value.\n"); pBit_flags = NULL; } } printf("Bit flags object memory print created!\n\n"); } else // Given number was 0 or negative { printf("Number of bits cannot be negative.\n"); pBit_flags = NULL; } return pBit_flags; } Status bit_flags_set_flag(BIT_FLAGS hBit_flags, int flag_position) { Status stat; int bit_to_set = flag_position; int* temp; Bit_flags* phBit_flags =…
- #ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…in c language !! typedef struct {long restaurant_id;char restaurant_name[10];char description[120];double rate;char cuisine[30]; opening_year;long capacity;char city[30];char address[60];char owner[30];} RECORD_t, *RECORD; RECdelete (unsigned long restaurant_id, RECORD *restaurant_array, unsigned long *p_size){int rec;int index;// index <- find the index of the RECORD with given restaurant_id in restaurant_array ????? (I DIDNT)if (index is valid)rec=restaurant_array[index];elserec=NULL;return rec;}When elements of only same data types are in one data structure, it is said to be O homogenous O LIFO O Unique O heterogenous
- В.width; } { t = B.type; w = { T.type = C.type; T.width = C.width; } T → B C В > int { B.type = integer; B.width 4; } В — foat { B.type = float; B.width = 8; } { C.type = t; C.width = w; } C - [ num] C1 = array(num. value, C1.type); { C.type C.width = num. value x C1. width; }typedef struct { short data[4];} MatrixElement; void copy_matrix(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[i*COLS+j].data[k]; } } }} void copy_matrix_transpose(MatrixElement m1[], MatrixElement m2[], int ROWS, int COLS) { int i, j, k; for (i = 0; i < ROWS; i++) { for (j = 0; j < COLS; j++) { for (k = 0; k < 4; k++) { m1[i*COLS+j].data[k] = m2[j*ROWS+i].data[k]; } } }} You can assume the following conditions: The matrix m1 is allocated at memory address 0, and matrix m2 immediately follows it. Indices i, j, and k are kept in registers. ROWS and COLS are constants. The cache is initially empty before the function call. The cache is write-back (i.e., only writes back to memory when a line is…Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…