Some simple C-language algorithms

1.

Enter a number to print the following graphic

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 graphic

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 graphic

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 graphic

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 ("A");

}

for (l=1;l<=num-i;l++)

{

printf ("B");

}

printf ("\ n");

}

return 0;

}

5.

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

#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 It's not a prime number to be able to do it.

}

}

if (i = = nu)//If equal to Nu proves that no factor is encountered

{

printf ("is prime number");

}

Else

{

printf ("Not prime number");

}

return 0;

}

6.

Enter two numbers for two-digit greatest common divisor (* * * * *)

The largest common factor, also known as greatest common divisor, is the largest 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);//Get I result is greatest common divisor

return 0;

}

7.

Enter two numbers for two-digit least common multiple (* * * *)

#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 the minimum number of just divisible two numbers is least common multiple

{

if (Max%numa = = 0 && Max%numb = = 0)

{

Break

}

max++;

}

printf ("%d", max);

return 0;

}

8.

Enter a number to decompose factorization (* * * * *)

Decomposition quality factor: the decomposition of a composite into a number of prime numbers product.

#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 to find the smallest factor (i)

printf ("%d", I);

Nu/=i; //Get

i--;

}

}

return 0;

}

9.

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

Outputs a value of 2 + 22 + 222. (Do not output the formula) (* * *)

M=5,n=a

3+33+333+3333+33333

10*3+3

10*33+3

Ret=10*ret+a

Sum=sum+ret

*/

#include <stdio.h>

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

Insert code here ...

int m,a;

scanf ("%d%d", &m,&a);

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 five digits, the symmetric number is called a palindrome number, and all the palindrome numbers are found.

such as 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.

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

Thought: Enter a number at a time to compare, 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; Tentative first number is maximum, very good thought

while (1)

{

scanf ("%d", &nu);

if (Nu = = 0)

{

Break

}

if (Max < NU)

{

Max=nu;

}

}

printf ("%d", max);

return 0;

}

12.

Enter a number to find the factorial of the 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.

Enter an integer (int range) number and output 10 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 every single one of these numbers and just put them in the right order.

while (1)

{

shu[i]=num1%10;

NUM1=NUM1/10;

i++;

if (NUM1 = = 0)

{

Break

}

}

Print out every bit of the number that's been put in order.

for (j=0;j<i;j++)

{

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

}

printf ("\ n");

return 0;

}

Some simple C-language algorithms