August, 2015 | Ajinkya Wavare

Conditional Statements

Problem Statement

if and else are two of the most heavily used conditionals in C/C++. They are used to execute zero or one statement among many statements.

They are be used in the following three ways.

if: It is used to execute a statement, given the condition is true.
```
if(condition) {
    ...
}
```
if – else: It is used to execute exactly one of the two statements.
```
if(first condition) {
    ...
}
else {
    ...
}
```

if – else if – else: It is used to execute one of the multiple statements.

if(first condition) {
    ...
}
else if(second condition) {
    ...
}
.
.
.
else if((n-1)'th condition) {

}
else {
    ...
}

You are given a positive integer, n,:

If 1≤n≤9, then print the English representation of it. That is “one” for 1, “two” for 2, and so on.
Otherwise print “Greater than 9” (without quotes).

Input Format

Input will contain only one integer, n.

Output Format

Print the output as described above.

Sample Input

Sample Output

five

Sample Input #01

Sample Output #01

eight

Sample Input #02

Sample Output #02

Greater than 9

Solution


#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n;
cin>>n;
if(n>9)
cout<<"Greater than 9"<<endl;
else if(n==9)
cout<<"nine"<<endl;
else if(n==1)
cout<<"one"<<endl;
else if(n==2)
cout<<"two"<<endl;
else if(n==3)
cout<<"three"<<endl;
else if(n==4)
cout<<"four"<<endl;
else if(n==5)
cout<<"five"<<endl;
else if(n==6)
cout<<"six"<<endl;
else if(n==7)
cout<<"seven"<<endl;
else
cout<<"eight"<<endl;

return 0;
}

A pointer in C is a way to share a memory address among different contexts (primarily functions). They are primarily used whenever a function needs to modify the content of a variable, of which it doesn’t have ownership.

In order to access the memory address of a variable, val, we need to prepend it with & sign. E.g., ‘‘&val“ returns the memory address of val.

This memory address is assigned to a pointer and can be shared among various functions. E.g. int∗p=&val will assign the memory address of val to pointer p. To access the content of the memory to which the pointer points, prepend it with a ‘‘∗“. For example, ∗p will return the value reflected by val and any modification to it will be reflected at the source (val).

void increment(int *v) {
    (*v)++;
}

int main() {
    int a;
    scanf("%d", &a);
    increment(&a);
    printf("%d", a);
    return 0;
}

You have to complete the function void update(int *a,int *b), which reads two integers as argument, and sets a with the sum of them, and b with the absolute difference of them.

a′=a+b
b′=|a−b|

Input Format

Input will contain two integers, a and b, separated by a newline.

Output Format

You have to print the updated value of a and b, on two different lines.
P.S.: Input/ouput will be automatically handled. You only have to complete the void update(int *a,int *b) function.

Sample Input

4
5

Sample Output

9
1

Explanation

a′=4+5=9
b′=|4−5|=1

Solution

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

void update(int *a,int *b) {
    // Complete this function 
    *a = *a+*b;
    *b = abs(*a-*b-*b);
}

int main() {
    int a, b;
    int *pa = &a, *pb = &b;
    
    scanf("%d %d", &a, &b);
    update(pa, pb);
    printf("%d\n%d", a, b);

    return 0;
}

Average Population of Each Continent

Problem Statement

Given two tables, City and Country whose description are given below. Print the name of all continents (key: Country.Continent) along with the average City population rounded down to nearest integer.

City

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

Country

+----------------+-------------+
| Field          | Type        |
+----------------+-------------+
| Code           | char(3)     |
| Name           | char(52)    |
| Continent      | char(50)    |
| Region         | char(26)    |
| SurfaceArea    | float(10,2) |
| IndepYear      | smallint(6) |
| Population     | int(11)     |
| LifeExpectancy | float(3,1)  |
| GNP            | float(10,2) |
| GNPOld         | float(10,2) |
| LocalName      | char(45)    |
| GovernmentForm | char(45)    |
| HeadOfState    | char(60)    |
| Capital        | int(11)     |
| Code2          | char(2)     |
+----------------+-------------+

PS #1: City.CountryCode and Country.Code is same key.
PS #2: Continent without cities should not be included in output.

Solution

  
Select Country.Continent,FLOOR(avg(City.Population)) from City Inner Join Country ON City.CountryCode = Country.Code Group by Country.Continent;

African Cities

Problem Statement

Given two tables, City and Country , whose descriptions are given below, you have to list the Name of all cities that are in the continent “Africa”.

City

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

Country

+----------------+-------------+
| Field          | Type        |
+----------------+-------------+
| Code           | char(3)     |
| Name           | char(52)    |
| Continent      | char(50)    |
| Region         | char(26)    |
| SurfaceArea    | float(10,2) |
| IndepYear      | smallint(6) |
| Population     | int(11)     |
| LifeExpectancy | float(3,1)  |
| GNP            | float(10,2) |
| GNPOld         | float(10,2) |
| LocalName      | char(45)    |
| GovernmentForm | char(45)    |
| HeadOfState    | char(60)    |
| Capital        | int(11)     |
| Code2          | char(2)     |
+----------------+-------------+

PS: City.CountryCode and Country.Code is same key.

Solution

  
Select City.Name from City Inner Join Country ON City.CountryCode = Country.Code and Continent='Africa';

Asian Population

Problem Statement

Given two tables, City and Country whose descriptions are given below. Find the sum of population of all the Cities that lies in “Asia” continent.

City

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

Country

+----------------+-------------+
| Field          | Type        |
+----------------+-------------+
| Code           | char(3)     |
| Name           | char(52)    |
| Continent      | char(50)    |
| Region         | char(26)    |
| SurfaceArea    | float(10,2) |
| IndepYear      | smallint(6) |
| Population     | int(11)     |
| LifeExpectancy | float(3,1)  |
| GNP            | float(10,2) |
| GNPOld         | float(10,2) |
| LocalName      | char(45)    |
| GovernmentForm | char(45)    |
| HeadOfState    | char(60)    |
| Capital        | int(11)     |
| Code2          | char(2)     |
+----------------+-------------+

PS: City.CountryCode and Country.Code is same key.

Solution

  
Select sum(City.Population) from City Inner Join Country ON City.CountryCode = Country.Code and Continent='Asia';

Population Density Difference

Problem Statement

Given a City table, whose fields are described as

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

print the difference between the maximum and minimum city populations.

Solution

  
select max(Population)-min(Population) from City;

Japan Population

Problem Statement

Given a City table, whose fields are described as

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

you have to print the sum of population of all the cities of Japan. The CountryCode for Japan is “JPN”.

Solution

  
select sum(Population) from City where CountryCode= 'JPN';

Average Population

Problem Statement

Given a City table, whose fields are described as

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

you have to print the average population of all cities, rounded down to the nearest integer.

Solution

  
select ROUND(avg(Population),0) from City;

Japanese Cities’ Name

Problem Statement

Given a City table, whose fields are described as

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

you have to print the name of all the cities of Japan. The CountryCode for Japan is “JPN”.

Solution

  
Select Name from City where CountryCode = 'JPN';

Japanese Cities’ Detail

Problem Statement

Given a City table, whose fields are described as

+-------------+----------+
| Field       | Type     |
+-------------+----------+
| ID          | int(11)  |
| Name        | char(35) |
| CountryCode | char(3)  |
| District    | char(20) |
| Population  | int(11)  |
+-------------+----------+

you have to print all the details of all the cities of Japan. The CountryCode for Japan is “JPN”.

Solution

  
Select * from City where CountryCode = 'JPN';

S	M	T	W	T	F	S
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31

Month: August 2015

Conditional Statements

Pointer

Average Population of Each Continent

African Cities

Asian Population

Population Density Difference

Japan Population

Average Population

Japanese Cities’ Name

Japanese Cities’ Detail