Several algorithms of C language seeking GCD and LCM __ algorithm

Find LCM algorithm:

LCM = product of two integers gcd

Find GCD algorithm:

(1) Euclidean method

There are two integers a and B:

①a%b to the remainder C

② if c=0, then B is two number of GCD

③ if c≠0, then a=b,b=c, then go back to execute ①

For example, the GCD process for 27 and 15 is:

27÷15 Yu 1215÷12 more than 312÷3 0 so, 3 is GCD

[CPP] view plain copy #include <stdio.h> void main ()/* Euclidean algorithm gcd/{int m, n, a, B, T,      C      printf ("Input two integer numbers:\n");      scanf ("%d%d", &a, &b);   M=a;      N=b;  while (b!=0)/* The remainder is not 0, continue dividing until the remainder is 0/{c=a%b; a=b;      B=c;}      printf ("The largest common divisor:%d\n", a);   printf ("The least common multiple:%d\n", m*n/a); }

⑵ Phase Subtraction

There are two integers a and B:

① if a>b, then A=a-b

② if a<b, then b=b-a

③ if A=b, then a (or b) is a two-digit GCD

④ if a≠b, then go back to execute ①

For example, the GCD process for 27 and 15 is:

27-15=12 (15>12) 15-12=3 (12>3)

12-3=9 (9>3) 9-3=6 (6>3)

6-3=3 (3==3)

Therefore, 3 is the GCD

[CPP]  View Plain  copy #include <stdio.h>   void main  ( )   /*  Phase subtraction for GCD  */   {        int m, n, a, b,  c;       printf ("input two integer numbers:\n");       scanf  ("%d,%d",  &a, &b); m=a; n=b;         /* a, b are not equal, and large numbers are reduced until they are equal. */        while  ( a!=b)               if  (a>b)   a=a-b;                  else   b=b-a;       printf ("the largest common divisor:%d\n",  a);      printf ("the least common multiple:%d\n",  m*n/a);  }&NBSP;&NBsp

⑶ method of poverty and lifting

There are two integers a and B:

①i=1

② if a,b can be divisible by I at the same time, then t=i

③i++

④ If I <= a (or b), then go back to execute ②

⑤ If i > A (or b), then T is GCD, end

Improved:

①i= A (or b)

② if a,b can be divisible by I at the same time, then I is GCD,

End

③i--, go back and execute ②.

There are two integers a and B:

①i=1

② if a,b can be divisible by I at the same time, then t=i

③i++

④ If I <= a (or b), then go back to execute ②

⑤ If i > A (or b), then T is GCD, end

Improved:

①i= A (or b)

② if a,b can be divisible by I at the same time, then I is GCD,

End

③i--, go back and execute ②.

[CPP]   View plain  copy #include <stdio.h>   void main  ()   /*  to seek GCD  */   {        int  m by exhaustive method, &NBSP;N,&NBSP;A,&NB