Num 19: Greatest common divisor. least common multiple. Judgment of Prime number

in C language learning, often encounter:

calculate greatest common divisor, least common multiple and prime number judgment problem;

To summarize here:

1. Greatest common divisor and least common multiple:

By mathematical knowledge we know:

least common multiple of two numbers = greatest common divisor of the product of these two numbers/two numbers;

So the problem of seeking greatest common divisor and least common multiple is actually a kind of problem;

①. Least common multiple:

Method One:

If x>y, from X, determine whether x can be divisible by y (x%y==0);

If divisible, x is both least common multiple;

If not divisible, calculate whether the 2x,3x,......nx can be divisible;

Implementation code:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #include <math.h> #include <string.h>int lcm (int x,int y) {int temp;if (y>x) { Temp=x;x=y;y=temp;} Temp=x;while (x%y!=0) X+=temp;return x;} int main () {int a,b,num;while (scanf ("%d%d", &a,&b)!=eof) {printf ("%d\n", (LCM (b)));} return 0;} </span>

Method Two:

Calculate the least common multiple indirectly by calculating the greatest common divisor:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #include <math.h> #include <string.h>int gcd (int x,int y) {int num=0;num= (x> Y?Y:X); while ((x%num!=0| | y%num!=0) &&num!=1) Num--;return num;} int main () {int a,b;while (scanf ("%d%d", &a,&b)!=eof) {printf ("%d\n", (a*b)/GCD (A, b));} return 0;} </span>

②. Greatest common divisor:

Method One:

(X<y) n determines whether or not X, Y is divisible by n at the same time.

If can x for greatest common divisor, otherwise judge n--;

See least common multiple code;

Method Two:

The Euclidean algorithm is simulated in a recursive way:

That is: X is the same as x%y (not 0) greatest common divisor;

Implementation code:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #include <math.h> #include <string.h>int gcd (int a,int b) {   if (b==0) return A;   return gcd (B,A%B);//No need to judge A, b size, has been automatically judged;} int main () {int a,b;while (scanf ("%d%d", &a,&b)!=eof) {printf ("%d\n", (GCD (b)));} return 0;} </span>

second, the prime number of the judgment:

Method One:

Judging one by one, from 2 to sqrt (x) One by one, x can be divisible;

If possible, for composite; otherwise, for prime numbers;

Implementation code:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #include <math.h>int sushu (int x) {for (int i=2;i<=sqrt (x); i++) if (x%i==0) return 0 ; return 1;} int main () {int n;while (scanf ("%d", &n)!=eof) if (Sushu (n)) printf ("yes\n"), Else printf ("no\n"); return 0;} </span>

Method Two:

The prime number plays the table judgment;

Create an array (only 0, 1 in the array) [1 for composite, 0 for prime number];

Starting from 2, backwards, all multiples of 2 must be composite, marked as 1;

Starting from 3, repeat the process, if some have been marked as composite, then skip;

Implementation code:

<span style= "FONT-SIZE:14PX;" > #include <stdio.h> #include <math.h> #include <stdlib.h> #include <string.h>int sushu[ 2000001];void sheet () {sushu[1]=0;sushu[2]=0;for (int i=2;i<=2000000;i++)        if (!sushu[i]) for        (int j=i+i;j <=2000000;j+=i)        sushu[j]=1;} int main () {    int n;memset (sushu,0,sizeof (Sushu));    Sheet ();    while (scanf ("%d", &n), N)    {        if (!sushu[n]) printf ("yes\n");        else printf ("no\n");    }    return 0;} </span>

All of the above are the basis of C language, only grasp the good foundation, can solve more profound problems!

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Num 19: Greatest common divisor. least common multiple. Judgment of Prime number