Basic C language examination questions

Basic C language examination questions

//

// Main. m

// C Language Test Library

//

// Created by MRBean on 15/5/20.

// Copyright (c) 2015 MRBean. All rights reserved.

//

 

# Import

 

/** 5, 12 questions

* Determine whether a number is a prime number.

*

* @ Param x indicates the number to be determined.

*

* @ Return 0 indicates that the return value is not a prime number, and 1 indicates a prime number.

*/

Int isZhiShu (int x)

{

Int count = 0;

For (int I = 1; I <= x; I ++ ){

If (x % I = 0 ){

Count ++;

}

}

If (count = 2 ){

Return 1;

}

Return 0;

}

 

// ================================================ ======================================

 

/**

* 6.

* Write a function to calculate the Triangle Area.

*/

Typedef struct SanJiaoXing

{

Double a; // the length of an edge of a triangle

Double B;

Double c;

 

} SanJiaoXing; // defines a struct variable. The variable type is SanJiaoXing.

 

Double mianJi (SanJiaoXing s) // 6th calculate the area of the triangle

{

 

Double area = 0; // area of the triangle

If (s. a + s. b> s. c & s. a + s. c> s. B & s. B + s. c> s. a) // The condition for forming a triangle is that the sum of any two sides is greater than the third side.

{

Double r = (s. a + s. B + s. c)/2;

Area = sqrt (r * (r-s.a) * (r-s. B) * (r-s.c); // use the Helen formula to calculate the area

Printf ("the Triangle area is % g", area );

Return area;

}

Else

{

Printf ("cannot form a triangle! ");

Return area;

 

}

 

 

}

// ================================================ ==========

 

/**

* 7. Determine whether the circle overlaps

*

*/

 

// Define a circular struct

Typedef struct Circle

{

Double x; // x coordinate of the center

Double y; // y coordinate of the center

Double r; // radius of the circle

} Circle;

 

Int chongDie (Circle a, Circle B)

{

Double distance = sqrt (pow (a. x-b.x, 2) + pow (a. y-b.y, 2); // distance between Centers

Double radiusSum = a. r + B. r;

If (distance

{

Printf ("two circles overlap! \ N ");

Return 1; // overlap

}

Else {

Printf ("two circles do not overlap! \ N ");

Return 0; // No overlap

 

}

}

 

// ==============================================

/**

* 8. String Encryption

*

*/

 

Char * JiaMi (char str [])

{

For (int I = 0; str [I]; I ++ ){

If (str [I] = 'Z '){

Str [I] = 'a ';

}

Else if (str [I] = 'Z ')

{

Str [I] = 'Z ';

}

Else if (str [I]> = 'A' & str [I] <= 'y ') | (str [I]> = 'A' & str [I] <= 'y '))

{

Str [I] ++;

}

 

}

Printf ("the encrypted string is % s", str );

Return str;

}

 

// ================================================ ==========

/**

* 11. define a three vertex (using a struct) to determine whether the three points can form a triangle. If a triangle can be formed, calculate the area of the triangle (Helen formula ); otherwise, print "cannot constitute a triangle ".

*

*/

 

// Define a point struct type

Typedef struct

{

Double x;

Double y;

} MPoint;

 

// Calculate the area of a triangle based on the three vertices.

Double mianJi2 (MPoint a, MPoint B, MPoint c)

{

Double len1 = sqrt (pow (a. x-b.x, 2) + pow (a. y-b.y, 2); // The distance between the AB

Double len2 = sqrt (pow (a. x-c.x, 2) + pow (a. y-c.y, 2); // distance between the ac

Double len3 = sqrt (pow (c. x-b.x, 2) + pow (c. y-b.y, 2); // The distance between cb

Double area = 0; // area of the triangle

If (len1 + len2> len3 & len1 + len3> len2 & len2 + len3> len1)

{

Double r = (len1 + len2 + len3)/2;

Area = sqrt (r * (r-len1) * (r-len2) * (r-len3 ));

Printf ("the Triangle area is % g", area );

Return area;

}

 

Else

{

Printf ("these three points cannot constitute a triangle! ");

Return area;

}

 

}

 

 

// ================================================ ==============

 

 

 

 

 

 

 

 

 

Int main (int argc, const char * argv []) {

 

// ================================================ ====

/**

* 1. Enter two positive integers m and n. Output the maximum public approx. And the minimum public multiples.

* Calculate the maximum common divisor c of two numbers a and B using the moving phase division.

*/

 

// Int m;

// Int n;

// Printf ("enter two numbers: \ n ");

// Scanf ("% d", & m, & n );

// Int temp = m * n; // store the product of m * n to calculate the minimum public multiple.

// Int c = m % n;

// While (c! = 0 ){

// M = n; // Divisor

// N = c; // the remainder is used as the divisor.

// C = m % n;

//}

// Printf ("maximum common divisor: % d, minimum common multiple % d \ n", n, temp/n );

 

// ================================================ ==========

/**

* 2. Enter a string to count the numbers of uppercase and lowercase letters.

*/

 

// Char str [1000];

// Int count1 = 0; // number of lower case Cases

// Int count2 = 0; // uppercase

// Printf ("enter a string \ n ");

// Gets (str );

// For (int I = 0; str [I]! = '\ 0'; I ++ ){

// If (str [I]> = 'A' & str [I] <= 'Z '){

// Count1 ++;

//}

// Else if (str [I]> = 'A' & str [I] <= 'Z '){

// Count2 ++;

//}

//}

// Printf ("the number of uppercase letters is % d, and the number of lowercase letters is % d", count1, count2 );

 

// ================================================ ============

 

/**

* 3. Enter two positive integers. program a and program n output the values of a + aa + aaa +... (a... a) (n ).

*/

 

// Printf ("enter two integers \ n ");

// Int;

// Int n;

// Int fac = 0; // expression in the form of a, aa, aaa...

// Int sum = 0; // calculate and

// Scanf ("% d", & a, & n );

// For (int I = 0; I

//{

// Fac = fac * 10 + a; // aa = a * 10 + a, and the last one equals the previous number * 10 +

// Sum + = fac;

//

//}

// Printf ("a + aa + aaa... = % d \ n", sum );

 

// ================================================ ========================

 

 

/**

* 4. Enter a positive integer n, and return 1! + 2! +... N! Value (n! Represents the factorial of n)

*/

 

// Int sum = 0; // The sum of the recorded factorial

// Int fac = 1; // records the factorial of a number.

// Int n; // Number Limit

// Printf ("enter an integer not greater than 10 \ n ");

// Scanf ("% d", & n );

// For (int I = 1; I <= n; I ++ ){

// Fac * = I;

// Sum + = fac;

//}

//

// Printf ("1! + 2! +... + N! = % D ", sum );

 

 

// ================================================ ================

 

/**

* 5. Write a function to determine the prime number, input an integer in the main function, and output the information about whether it is a prime number.

*/

 

// Int n;

// Printf ("enter an integer ");

// Scanf ("% d", & n );

// If (isZhiShu (n) = 1 ){

// Printf ("is a prime number! \ N ");

//}

// Else

//{

// Printf ("not a prime number! ");

//}

 

// ================================================ ================

 

 

/**

* 6. Define a struct that contains three sides of a triangle. Write a function, input the struct variable, and return the Triangle Area (Note whether the triangle is valid)

*/

 

// SanJiaoXing s;

// Printf ("Enter the three sides of the triangle \ n ");

// Scanf ("% lf", & s. a, & s. B, & s. c );

// MianJi (s); // calculate the Triangle Area

//

 

 

// ================================================ ======================================

 

 

/**

* 7. Define a struct that contains the center and radius. Write a function, input two struct variables, and return whether the two circles overlap.

*/

 

 

// Circle a; // The First Circle

// Circle B; // The second Circle

// Printf ("Enter the center and radius of the first circle \ n ");

// Scanf ("% lf", & a. x, & a. y, & a. r );

// Printf ("Enter the center and radius of the second circle \ n ");

// Scanf ("% lf", & B. x, & B. y, & B. r );

// ChongDie (a, B );

 

// ================================================ ======================================

 

/**

* 8. Write a function to encrypt the string. The encryption rule is a-> B z-> a and so on.

*/

 

// Char s [1000];

// Gets (s );

// JiaMi (s );

 

// ================================================ ======================================

 

/**

* 9. If you enter 10 numbers, the maximum and minimum values are returned.

*/

 

// Int a [10];

// Int max = 0;

// Int min = 0;

// Printf ("input 10 numbers \ n ");

// For (int I = 0; I <10; I ++)

//{

// Scanf ("% d", & a [I]);

// If (I = 0)

//{

// Max = a [0];

// Min = a [0];

//}

//

// If (max

//{

// Max = a [I];

//}

//

// If (min> a [I])

//{

// Min = a [I];

//}

//

//}

// Printf ("maximum value: % d \ n minimum value: % d \ n", max, min );

//

// Printf ("Enter ten \ n ");

// Int a [10];

// For (int I = 0; I <10; I ++ ){

// Scanf ("% d", & a [I]);

//}

// Int max = a [0]; // maximum value

// Int min = a [0]; // minimum value

// For (int I = 1; I <10; I ++ ){

// If (max

// Max = a [I];

//}

// If (min> a [I]) {

// Min = a [I];

//}

//}

//

// Printf ("maximum value is % d, minimum value is % d", max, min );

//

// ================================================ ==================================

 

/**

*

 

10. The following code has problems. Please point out

 

Int * p;

{

Int a; // a is released after parentheses.

P = &;

}

* P = 3; // The problem is that the wild pointer is abnormal.

 

*/

// ================================================ ==================================

 

/**

* 11. define a three vertex (using a struct) to determine whether the three points can form a triangle. If a triangle can be formed, calculate the area of the triangle (Helen formula ); otherwise, print "cannot constitute a triangle ".

*/

 

// MPoint a; // vertex of the triangle

// MPoint B;

// MPoint c;

// Printf ("Enter three vertices in sequence! \ N ");

// Scanf ("% lf", &. x, &. y, & B. x, & B. y, & c. x, & c. y );

// MianJi2 (a, B, c );

 

 

// ================================================ ==================================

 

/**

* 12. Print up to 100 prime numbers

*/

 

// Int count = 0; // number of printed prime numbers

// For (int I = 1; count <= 100; I ++) // The cycle condition is in count <= 100, that is, if no 100 prime numbers are found, the loop goes on.

//{

// If (isZhiShu (I) = 1) {// a prime number is encountered.

// Printf ("% d", I );

// Count ++;

//}

//

//}

 

// ================================================ ==================================

 

/**

* 13. Enter 10 numbers and calculate the average value.

*/

 

// Double a [10]; // used to save 10 numbers

// Double avg; // Average Value

// Double sum = 0.0;

// Printf ("Enter ten \ n ");

// For (int I = 0; I <10; I ++)

//{

// Scanf ("% lf", & a [I]);

// Sum + = a [I];

//}

//

// Avg = sum/10;

// Printf ("average value: % g", avg );

 

// ================================================ ================

 

/**

* 14. Calculate the result of 1-2 + 3-4 + 5-6 +... 100

*/

 

// Int sum = 0;

// For (int I = 1; I <= 100; I ++)

//{

// If (I % 2 = 0)

//{

// Sum-= I;

//}

// Else

//{

// Sum + = I;

//}

//}

//

// Printf ("and % d", sum );

 

// ================================================ ========================

 

 

/**

* 15. The input positive integer n program outputs the value of 1/2 + 2/3 +... n/n + 1.

*/

 

 

// Int sum = 0;

// Int n;

// Printf ("enter a positive integer! \ N ");

// Scanf ("% d", & n );

// For (int I = 1; I <= n; I ++)

//{

// Sum + = I/(I + 1 );

//}

// Printf ("the sum of calculation is % d", sum );

 

// ================================================ ==================================

Return 0;

}