求两个整数的最大公约数和最小公倍数
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intfun(inta,intb){inti,j,m,n;if(a>b){m=a;a=b;b=m;}i=a;j=b;while((n=j%i)!=0){j=i;i=n;}returni;}
privatesubform_click()n=val(text1)m=val(text2)ifn
packagecom.fmzrt;/***求两个数的最大公约数和最小公倍数*@authorkele**/publicclassGongyueGongbeiShu{/***求两个数的最大公约数*@par
#includeintgongyue(intm,intn){intr;if(m==n)returnm;elsewhile((r=m%n)!=0){m=n;n=r;}returnn;}intgongbe
#includeintgongyue(intm,intn){intr;if(m==n)returnm;elsewhile((r=m%n)!=0){m=n;n=r;}returnn;}voidmain(
#includeintyue(inta,intb){inti,y;for(i=b;i>=1;i--)if(a%i==0&&b%i==0){y=i;break;}returny;}intbei(inta
PrivateSubCommand1_Click()DimSAsLongS=YueOrBei(24,32)MsgBox"24和32的最大公约数是:"&S,vbInformationS=YueOrBei
intmax_common_divisor(inta,intb){//最大公约数intlarge_num,small_num,r;if(a>b){large_num=a;small_num=b;}el
#includevoidmain(){inthcf(int,int);intlcd(int,int,int);intu,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);pr
C语言很久没写了,语法都忘了.给你个思路,假设两个数中n比较小,其实无所谓,只不过循环次数少一点.ints1=1;//假设最大公约数是1.for(inti=1;i
一般思维:#includeusingnamespacestd;intmain(){intx,y,max,min,k;inttem0,tem1;couty;max=x>y?x:y;min=x
EXCELVBA求公约数函数过程如下:PublicFunctionGYS(MAsInteger,NAsInteger)AsIntegerDimYS,YS1,MX,MNMX=WorksheetFunct
PrivateSubForm_Load()DimXAsLong,YAsLong,KAsLongDimAAsLong,BAsLongA=6B=8X=IIf(A>=B,A,B)'x存入最大值Y=IIf(A
例子一:#includevoidmain(){voidz1(inta,intb);intx,y,temp;scanf("%d%d",&x,&y);if(x
这种方法是数学里面的辗转相除法.具体思路为:假设a=15,b=9那么r=a%b则r=15%9=6,由于余数不为0,所有9不是最大公约数现在令a=b,b=r,那么a=9,b=6,继续求r=a%b则r=9
#include"stdio.h"voidmain(){\x05intnum1,num2,temp,a;\x05printf("pleaseinputtwonumbers:\n");\x05scanf
占天时地利人和取九州四海财宝横批:财源不断
//输入的两个数请用空格分隔,如:1218#includeintmain(){intp,r,n,m,temp;printf("请输入两个正整数n,m:");scanf("%d%d,",&n,&m);i
①#includeinthcf(inta,intb){\x09intc;\x09while(b){c=b;b=a%b;a=c;}\x09returna;}intlcd(inta,intb,intc){
完整程序如下:#includefun(intx,inty){intr;if(x>y){x=x;y=y;}r=x;x=y;y=r;r=x%y;while(r!=0){x=y;y=r;r=x%y;}ret