3.1 Greatest common divisor and least common multiple, with global variables and functions

Method One: Poor lifting

#include <stdio.h>intMax, Min;intMain () {intMaxintAintb);//can be declared as void because the return value is not required.     intMinintAintb); intA, B; scanf ("%d%d", &a, &b); printf ("\ n"); Max (A, b);//The function can be called directly here, if the global variable assignment will change to 0. Min (A, b);//that is, if you have max = Max (A, A, a, b), you will cause Max to change its value to 0. printf"Max is%d and Min is%d\n", Max, Min); return 0; } intMaxintXinty) {intI, C;  for(i = x > y y:x; i >0; i--){         if(x% i = =0&& y% i = =0) {Max=i;  Break; }     }     return 0;//since it is a global variable, you do not need to return the value of this sentence can be 0 or not write.       } intMinintXintY//I am not using a division, but a poor lift.  {     intI, C;  for(i = x > y x:y; i >0; i++){         if(i% x = =0&& I% y = =0) {Min=i;  Break; }     }     return 0; }

Method Two: Divide

#include <stdio.h>intHCF,LCD;intMain () {voidHcfint,int);//because only the function is borrowed, the global variable is not required to return the value    voidLcdint,int); intu, v; scanf ("%d,%d", &u, &v);    HCF (U,V);    LCD (u,v); printf ("h.c.f =%d\n", HCF);//Remember, two global variables are nested functions, such as the nesting of printfprintf"l.c.d =%d\n", LCD);//Otherwise, the value is wrong    return 0; } voidHcfintUintV//Euclidean Method {     intT, R; if(V >u) {T= u; U = V; v =T; }      while((r = u% v)! =0) {u=v; V=R; } HCF=v;} voidLcdintUintv) {LCD= U * V/HCF;}

From study Counseling

3.1 Greatest common divisor and least common multiple, with global variables and functions