用主函数调用求2个数的最大公约数和最小公倍数

1循环结构A=rand(1,20)Amin=A(1);Amax=A(1);fori=1:19ifA(i+1)>AmaxAmax=A(i+1);endifA(i+1)

#include<stdio.h>void sort(float *a, int n){ int i,j,tmp; &

whilesqrt((b(:,k)-a(:,k))'*(b(:,k)-a(:,k)))>=apxlFLK=fun(lbt(:,k));%%%%%调用函数funFMK=fun(mu(:,k));%%%%

#include"stdio.h"voidmax_min(inta[],intn,int*max,int*min){int*p;*max=*min=*a;for(p=a+1;pif(*p>*max)*

#include#definen5intgetmax(inta[]){inti,max;max=a[0];for(i=0.;imax)max=a[i];}returnmax;}intmain(void

用递归函数#include#includeintinv(intx);main(){intsum=0,i;for(i=1;i

#include<stdio.h>int getmax(int a,int b){      re

用循环语句把20个数组成数组A,执行下面的程序,可以直接得到最大和最小值.A=[123456723321253123535235315131332321];max=A(1);min=A(1);fori

#includeintsushu(intn){intj;for(j=2;j

intmax(inta,intb)//比较大数,返回值为大数{if(a>b){returna;}else{returnb;}}voidmain(){intx=9;inty=5;intz=max(x,y

#includevoidmain(){inthcf(int,int);intlcd(int,int,int);intu,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);pr

PrivateSubCommand1_Click()DimIAsIntegerDimN(15)AsInteger,MAsIntegerRandomizeForI=1To15N(I)=Int(Rnd*1

==||童鞋.你是要返回输入的n个数的最大值,还是一个数组的最大值啊.返回输入n个数最大值的函数：#includeintMax(intn);//最大值函数声明intmain(){intn;scanf(

#includefloatmax(floata,floatb){if(a>b)returna;returnb;}voidmain(){floata,b,c,d;scanf("%f,%f,%f",&a,

3个

#include"stdio.h"intfunc(inta,intb){inttemp;while(b!=0)/*利用辗除法,直到b为0为止*/{temp=a%b;a=b;b=temp;}retu

#include#includeusingnamespacestd;intjust(intnum){for(inti=2;i

#include#includevoidb1(){floatl,s,k;inta,b,c,h;printf("\n");printf("\n");printf("输入a,b,c的值\n");print

#includeintmaxy(int&a,int&b){intn,i,j;for(i=1;i>x>>y;s=maxy(x,y);cout

#includevoidswap(int*x,int*y);intmain(void){inta,b;printf("a:");scanf("%d",&a);printf("b:");scanf("%