3.1 Conditional Statement

কন্ডিশনাল স্টেটমেন্ট (৪)

1. শূন্য, ধনাত্মক, ঋণাত্মক - সংখ্যার চিহ্ন নির্ণয়

   1. [1.1] একটি পূর্ণসংখ্যা শূন্য কিনা নির্ণয় কর। (Similar)

   2. [1.2] একটি পূর্ণসংখ্যা ধনাত্মক না ঋণাত্মক নির্ণয় কর। (Similar)

   3. [1] একটি পূর্ণসংখ্যা শূন্য না ধনাত্মক না ঋণাত্মক নির্ণয় কর।

2. [2] বয়স অনুযায়ী একজন ব্যাক্তি ভোটার কিনা নির্ণয় কর।

3. [3] বয়স অনুযায়ী চাকরির আবেদন করার যোগ্য কিনা নির্ণয় কর। (নির্ধারিত বয়সসীমা ১৮ থেকে ৩৫)

4. [4] দুইটি জটিল সংখ্যা সমান কিনা নির্ণয় কর।

সমাধান

[1.1] একটি পূর্ণসংখ্যা শূন্য কিনা নির্ণয় কর।

Partial Solution

#include<stdio.h>
int main(){
    int a = 0; // Output => Zero
    //int a = 1; // No Output
    
    if( a == 0 ){ // Testing Equlity Operator [==]
        printf("Zero\n");
    }
    
    return 0;
}

only if solution

#include<stdio.h>
int main(){
    int a = 0; // Output => Zero
    //int a = 1; // Output => Not Zero
    
    if( a == 0 ){ // Testing Equlity Operator [==]
        printf("Zero\n");
    }
    if( a != 0 ){ // Testing Not Equlity Operator [!=]
        printf("Not Zero\n");
    }
    
    return 0;
}

if else solution

#include<stdio.h>
int main(){
    int a = 0; // Output => Zero
    //int a = 1; // Output => Not Zero
    
    if( a == 0 ){
        printf("Zero\n");
    }
    else{ // Else block will execute when (a == 0) is false
        printf("Not Zero\n");
    }
    
    return 0;
}

advanced solution

#include<stdio.h>
int main(){
    int a = 0; // Output => Zero
    //int a = 1; // Output => Not Zero
    
    if( a ){ // if evaluate 0 as FALSE and non-zero value as TRUE
    // This block will execute when the value of a is non-zero
        printf("Not Zero\n");
    }
    else{
    // This block will execute when the value of a is 0
        printf("Zero\n");
    }
    
    return 0;
}

more advanced

#include<stdio.h>
int main(){
    int a = 0; // Output => Zero
    //int a = 1; // Output => Not Zero
    
    // Logical NOT Operator
    if( !a ){ // when a=0, it means a is FALSE, !a means TRUE
    // This block will execute when the value of a is 0
        printf("Zero\n");
    }
    else{
    // This block will execute when the value of a is non-zero
        printf("Not Zero\n");
    }
    
    return 0;
}

[1.2] একটি পূর্ণসংখ্যা ধনাত্মক না ঋণাত্মক নির্ণয় কর।

only if solution

#include<stdio.h>
int main(){
    int a = 5; // Output => Positive
    //int a = -5; // Output => Negative
    //int a = 0; // No Output
    
    if( a > 0 ){ // Testing Grater Then Operator [>]
        printf("Positive\n");
    }
    if( a < 0 ){ // Testing Less Then Operator [<]
        printf("Negative\n");
    }
    
    return 0;
}

if else solution

#include<stdio.h>
int main(){
    int a = 5; // Output => Positive
    //int a = -5; // Output => Negative
    //int a = 0; // Output => Negative
    
    if( a > 0 ){ // Testing Grater Then Operator [>]
        printf("Positive\n");
    }
    else{ // Else block will execute when (a > 0) is false
        printf("Negative\n");
    }
    
    return 0;
}

next solution

#include<stdio.h>
int main(){
    int a = 5; // Output => Positive
    //int a = -5; // Output => Negative
    //int a = 0; // Output => Positive
    
    if( a >= 0 ){ // Testing Grater Then or Equal Operator [>=]
        printf("Positive\n");
        // Zero is neither Position nor Negative
        // So This Solution is not perfect
    }
    else{ // Else block will execute when (a > 0) is false
        printf("Negative\n");
    }
    
    return 0;
}

[1] একটি পূর্ণসংখ্যা শূন্য না ধনাত্মক না ঋণাত্মক নির্ণয় কর।

only if solution

#include<stdio.h>
int main(){
    int num = 5; // Output => Positive
    //int num = -5; // Output => Negative
    //int num = 0; // Output => Zero
    
    if( num > 0 ){ // Testing Grater Then Operator [>]
        printf("Positive\n");
    }
    if( num < 0 ){ // Testing Less Then Operator [<]
        printf("Negative\n");
    }
    if( num == 0 ){ // Testing Equlity Operator [==]
        printf("Zero\n");
    }
    
    return 0;
}

if else if solution

#include<stdio.h>
int main(){
    int num = 5; // Output => Positive
    //int num = -5; // Output => Negative
    //int num = 0; // Output => Zero
    
    if( num > 0 ){ // Testing Grater Then Operator [>]
        printf("Positive\n");
    }
    else if( num < 0 ){ // Testing Less Then Operator [<]
        printf("Negative\n");
    }
    else if( num == 0 ){ // Testing Equlity Operator [==]
        printf("Zero\n");
    }
    
    return 0;
}

if else if ladder solution

#include<stdio.h>
int main(){
    int num = 5; // Output => Positive
    //int num = -5; // Output => Negative
    //int num = 0; // Output => Zero
    
    if( num > 0 ){ // Testing Grater Then Operator [>]
        printf("Positive\n");
    }
    else if( num < 0 ){ // Testing Less Then Operator [<]
        printf("Negative\n");
    }
    else{
    // This else block will execute when both previous conditions are false
        printf("Zero\n");
    }
    
    return 0;
}

Nested Connditional Statement

#include<stdio.h>
int main(){
    int num = 5; // Output => Positive
    //int num = -5; // Output => Negative
    //int num = 0; // Output => Zero
    
    if( num == 0 ){ // Testing Equlity Operator [==]
        printf("Zero\n");
    }
    else{
        if( num > 0 ){ // Testing Grater Then Operator [>]
            printf("Positive\n");
        }
        //else if( num < 0 ){ // Testing Less Then Operator [<]
        else{
            printf("Negative\n");
        }
    }
    
    return 0;
}

Sample Input	Sample Output
5	Positive
-3	Negative
0	Zero

[2] বয়স অনুযায়ী একজন ব্যাক্তি ভোটার কিনা নির্ণয় কর।

Test Voting Eligibility 1 (only if solution)

Partial Solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // No Output
 
    if(age > 18){
        printf("Eligible to Vote\n");
    }
    if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

only if solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age == 18){
        printf("Eligible to Vote\n");
    }
    if(age > 18){
        printf("Eligible to Vote\n");
    }
    if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

Logical Operator (OR) Solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age == 18 || age > 18){
        printf("Eligible to Vote\n");
    }
    if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

better solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age >= 18){
        printf("Eligible to Vote\n");
    }
    if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

Test Voting Eligibility 2 (if else solution)

if else if solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age >= 18){
        printf("Eligible to Vote\n");
    }
    else if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

if else solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age >= 18){
        printf("Eligible to Vote\n");
    }
    else{
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

advanced solution

#include<stdio.h>
#define VOTE_AGE 18 // Define VOTE_AGE macro
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if( age >= VOTE_AGE ){
        printf("Eligible to Vote\n");
    }
    else{
        printf("Not Eligible to Vote\n");
    }
    
    return 0;
}

Test Voting Eligibility 3 (alternative solution)

alternative solution 1

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age < 18){
        printf("Not Eligible to Vote\n");
    }
    else{
        printf("Eligible to Vote\n");
    }
    
    return 0;
}

alternative solution 2

#include<stdio.h>
int main(){
    int age = 20; // Output => Voter
    //int age = 15; // Output => Not Voter
    //int age = 18; // Output => Voter
 
    if(age <= 17){
        printf("Not Eligible to Vote\n");
    }
    else{
        printf("Eligible to Vote\n");
    }
    
    return 0;
}

[3] বয়স অনুযায়ী চাকরির আবেদন করার যোগ্য কিনা নির্ণয় কর। (নির্ধারিত বয়সসীমা ১৮ থেকে ৩৫)

Nested Condition

#include<stdio.h>
int main(){
    int age = 20; // Output => Eligible to apply
    //int age = 15; // Output => Not Eligible to apply
    //int age = 38; // Output => Not Eligible to apply
 
    if(age >= 18){ // Testing Grater Then or Equal Operator [>=]
        if( age <= 35 ){ // Testing Less Then or Equal Operator [<=]
            printf("Eligible to apply\n");
        }
        else{
            printf("Not eligible to apply\n");
        }
    }
    else{
        printf("Not eligible to apply\n");
    }
    
    return 0;
}

Logical AND

#include<stdio.h>
int main(){
    int age = 20; // Output => Eligible to apply
    //int age = 15; // Output => Not Eligible to apply
    //int age = 38; // Output => Not Eligible to apply
 
    if(age >= 18 && age <= 35){ // Testing Logical AND Operator [&&]
    // Print "Eligible to apply" if the age is between 18 and 35 inclusive
        printf("Eligible to apply\n");
    }
    else{
    // Print "Not Eligible to apply" if the age is outside the range
        printf("Not eligible to apply\n");
    }
    
    return 0;
}

advanced solution

#include<stdio.h>
int main(){
    int age = 20; // Output => Eligible to apply
    //int age = 15; // Output => Not Eligible to apply
    //int age = 38; // Output => Not Eligible to apply
 
    if(age <= 18){ // Testing Less Then or Equal Operator [<=]
        printf("Not eligible to apply\n");
    }
    else if( age >= 35 ){ // Testing Grater Then or Equal Operator [>=]
        printf("Not eligible to apply\n");
    }
    else{
        printf("Eligible to apply\n");
    }
    
    return 0;
}

Logical OR

#include<stdio.h>
int main(){
    int age = 20; // Output => Eligible to apply
    //int age = 15; // Output => Not Eligible to apply
    //int age = 38; // Output => Not Eligible to apply
 
    if(age <= 18 || age >= 35){ // Testing Logical OR Operator [||]
        printf("Not eligible to apply\n");
    }
    else{
        printf("Eligible to apply\n");
    }
    
    return 0;
}

Sample Input	Sample Output
20	Eligible to apply
15	Not Eligible to apply
38	Not Eligible to apply

[4] দুইটি জটিল সংখ্যা সমান কিনা নির্ণয় কর।

Nested Conditional Statement

#include<stdio.h>
int main(){
    int a1 = 0, b1 = 1, a2 = 0, b2 = 1; // Output => Equal
    //int a1 = 1, b1 = 1, a2 = 0, b2 = 0; // Output => Not Equal
    
    if( a1 == a2 ){
        if( b1 == b2 ){
            printf("Equal\n");
        }
        else{
            printf("Not Equal\n");
        }
    }
    else{
        printf("Not Equal\n");
    }
    
    return 0;
}

Logical Operator (AND) Solution

#include<stdio.h>
int main(){
    int a1 = 0, a1 = 1, a2 = 0, b2 = 1; // Output => Equal
    //int a1 = 1, b1 = 1, a2 = 0, b2 = 0; // Output => Not Equal
    
    if( a1 == a2 && b1 == b2 ){
        printf("Equal\n");
    }
    else{
        printf("Not Equal\n");
    }
    
    return 0;
}

Sample Input	Sample Output
3 4 3 4	The complex numbers are equal
5 6 5 7	The complex numbers are not equal