Some simple C language algorithms and simple C language Algorithms

Some simple C language algorithms and simple C language Algorithms

1.

Enter a number to print the following figure.

Input 5

Print:

*

**

***

****

*****

# Include <stdio. h>

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

Int num, I, j;

Scanf ("% d", & num );

For (I = 1; I <= num; I ++)

{

For (j = 1; j <I + 1; j ++)

{

Printf ("*");

}

Printf ("\ n ");

}

Return 0;

}

 

2.

Enter a number to print the following figure.

Input 5

Print:

*

**

***

****

*****

# Include <stdio. h>

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

Int num, I, j, k;

Scanf ("% d", & num );

For (I = 1; I <= num; I ++)

{

For (j = 1; j <= num-I; j ++)

{

Printf ("");

}

For (k = 1; k <I + 1; k ++)

{

Printf ("*");

}

Printf ("\ n ");

}

Return 0;

}

 

3.

Enter a number to print the following figure.

Input 5

Print:

*

***

*****

*******

*********

# Include <stdio. h>

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

Int num, I, j, k, l;

Scanf ("% d", & num );

For (I = 1; I <= num; I ++)

{

For (j = 1; j <= num-I; j ++)

{

Printf ("");

}

For (k = 1; k <I * 2-1 + 1; k ++)

{

Printf ("*");

}

For (l = 1; l <= num-I; l ++)

{

Printf ("");

}

Printf ("\ n ");

}

Return 0;

}

 

4.

Enter a number to print the following figure.

Input 5

Print

ABBBB

AABBB

AAABB

AAAAB

AAAAA

# Include <stdio. h>

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

Int num, I, j, k, l;

Scanf ("% d", & num );

For (I = 1; I <= num; I ++)

{

For (j = 1; j <= num-I; j ++)

{

Printf ("");

}

For (k = 1; k <I + 1; k ++)

{

Printf ("");

}

For (l = 1; l <= num-I; l ++)

{

Printf ("B ");

}

Printf ("\ n ");

}

Return 0;

}

 

5.

Enter a number to determine whether it is a prime number )(****)

# Include <stdio. h>

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

Int nu, I;

Scanf ("% d", & nu );

For (I = 2; I <nu; I ++)

{

If (nu % I = 0)

{

Break; // if it can be fully divided, it indicates that it is not a prime number.

}

}

If (I = nu) // if it is equal to nu, it indicates that it is not a factor.

{

Printf ("is a prime number ");

}

Else

{

Printf ("not a prime number ");

}

Return 0;

}

 

6.

Enter two numbers and calculate the maximum common divisor of the two numbers (*****)

// The maximum public factor, also known as the maximum public approx. And the maximum public factor, refers to the maximum total number of two or more Integers

# Include <stdio. h>

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

Int I, numA, numB, mix;

Scanf ("% d, % d", & numA, & numB );

Mix = (numA <numb )? NumA: numB;

For (I = mix; I> = 1; I --)

{

If (numA % I = 0 & numB % I = 0)

{

Break;

}

}

Printf ("% d", I); // The result of I is the maximum public approx.

Return 0;

}

 

7.

Enter two numbers to obtain the minimum public multiple of them (****)

 

# Include <stdio. h>

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

Int numa, numb, max;

Scanf ("% d, % d", & numa, & numb );

Max = (numa <numb )? Numa: numb;

While (1) // find out that the smallest number that can just divide two numbers is the least common multiple.

{

If (max % numa = 0 & max % numb = 0)

{

Break;

}

Max ++;

}

Printf ("% d", max );

Return 0;

}

 

8.

Enter a number to separate the prime factor (*****)

Factorization factor: Splits a sum into the product of several prime numbers.

 

# Include <stdio. h>

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

Int nu;

Scanf ("% d", & nu );

For (int I = 2; I <= nu; I ++ ){

If (nu % I = 0) {// use the smallest factor (I)

Printf ("% d", I );

Nu/= I; // get

I --;

}

}

Return 0;

}

 

9.

Enter two numbers n, a, if n = 3, a = 2;

Output 2 + 22 + 222 values. (No output sub-statement is required )(****)

M = 5, n =

3 + 33 + 333 + 3333 + 33333

10*3 + 3

10*33 + 3

Ret = 10 * ret +

Sum = sum + ret

 

*/

# Include <stdio. h>

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

// Insert code here...

Int m,;

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

Int sum = 0;

Int ret = 0;

For (int I = 1; I <= m; I ++)

{

Ret = 10 * ret + a; // ret is the current value, such as 3,33, 333,3333

Sum = sum + ret;

}

Printf ("% d \ n", sum );

Return 0;

}

 

 

10.

In the five-digit number, the symmetric number is called the return number to find all the return numbers.

For example, 12321 (***)

 

# Include <stdio. h>

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

// Insert code here...

For (int I = 10000; I <= 99999; I ++)

{

Int bita, bitb, bitc, bitd;

// 12345% 10 = 5

Bita = I % 10;

// 12345/10 = 1234 1234% 10 = 4

Bitb = I/10% 10;

// 12345/1000 = 123 12% 10 = 2

Bitc = I/1000% 10;

// 12345/10000 = 1 1% 10 = 1

Bitd = I/10000% 10;

If (bita = bitd & bitb = bitc)

{

Printf ("% d", I );

}

}

Return 0;

}

 

 

11.

Input any majority, the last number is 0, and the maximum number of these numbers is output. (**)

Thought: input a number each time for comparison, and save the current maximum number each time.

# Include <stdio. h>

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

// Insert code here...

Int nu, max;

Scanf ("% d", & nu );

Max = nu; // The tentative first number is the maximum value, which is a good idea.

While (1)

{

Scanf ("% d", & nu );

If (nu = 0)

{

Break;

}

If (max <nu)

{

Max = nu;

}

}

Printf ("% d", max );

Return 0;

}

 

12.

Enter a number to calculate the factorial of this number. (*)

# Include <stdio. h>

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

Int num1, I, sum = 1;

Scanf ("% d", & num1 );

For (I = 1; I <= num1; I ++)

{

Sum * = I;

}

Printf ("% d \ n", sum );

Return 0;

}

 

13.

Input an integer (within the int range) and output 10 hexadecimal values in reverse order.

# Include <stdio. h>

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

Int num1, I = 0, shu [64], j;

Scanf ("% d", & num1 );

// Take out each digit of this number, and arrange the order right after it is obtained.

While (1)

{

Shu [I] = num1 % 10;

Num1 = num1/10;

I ++;

If (num1 = 0)

{

Break;

}

}

// Print each digit of the number in the order

For (j = 0; j <I; j ++)

{

Printf ("% d", shu [j]);

}

Printf ("\ n ");

Return 0;

}