Basic Data Types C++
Problem Statement :
Some C++ data types, their format specifiers, and their most common bit widths are as follows: Int ("%d"): 32 Bit integer Long ("%ld"): 64 bit integer Char ("%c"): Character type Float ("%f"): 32 bit real value Double ("%lf"): 64 bit real value Reading To read a data type, use the following syntax: scanf("`format_specifier`", &val) For example, to read a character followed by a double: char ch; double d; scanf("%c %lf", &ch, &d); For the moment, we can ignore the spacing between format specifiers. Printing To print a data type, use the following syntax: printf("`format_specifier`", val) For example, to print a character followed by a double: char ch = 'd'; double d = 234.432; printf("%c %lf", ch, d); Note: You can also use cin and cout instead of scanf and printf; however, if you are taking a million numbers as input and printing a million lines, it is faster to use scanf and printf. Input Format Input consists of the following space-separated values: int, long, char, float, and double, respectively. Output Format Print each element on a new line in the same order it was received as input. Note that the floating point value should be correct up to 3 decimal places and the double to 9 decimal places.
Solution :
Solution in C :
#include <iostream>
#include <cstdio>
using namespace std;
int main() {
// Complete the code.
int a,c;
long long b;
char d;
float f;
double e;
scanf("%d %d %lld %c %f %lf", &a ,&c,&b,&d,&f,&e);
printf("%0d\n",a);
printf("%0d\n",c);
printf("%0lld\n",b);
printf("%0c\n", d);
printf("%0f\n", f);
printf("%0lf\n", e);
return 0;
}
View More Similar Problems
Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →Reverse a doubly linked list
This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.
View Solution →Tree: Preorder Traversal
Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's
View Solution →Tree: Postorder Traversal
Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the
View Solution →