蓝桥杯编写一函数lcm,求两个正整数的最小公倍数.

#include#include#include/*利用辗转相除法求最大公约数*/intgcd(intn,intm){intr;if(n

#includemain(){inta,c,b,d;scanf("%d%d",a,b);c=a+b;d=a*b;printf("%d%d",c,d);}再问：采用函数的方法再答：先输入两个数，然后执行

这个简单：#includeusingnamespacestd;intHe(intx,inty){intz;z=x+y;returnz;}intCha(intx,inty){intz;z=x-y;ret

intmin(intx,inty){if(x>y)returny;returnx;}再问：能把主函数也写一下么再答：#includevoidmain(){intx,y;printf("请输入两个整数:

#includeintadd(inta,intb){returna+b;}main(){inta,b;scanf("%d%d",&a,&b);printf("a+b=%s",add(a,b));}

publicintmax(inta,intb){returna>b?a:b;}publicdoublemax(doublea,doubleb,doublec){doublet=a>b?a:b;retu

int function(double **p, int p_r,int p_c, double **q,int q_r

这道题目也不是很难,自家先做一下.思路很清晰,没有涉及到算法问题.

#includeintfun(intm,intn){inti,s=1;for(i=2;i

调试过了,如果需要小数,把int换成float就行了#include"stdio.h"intSub(inta,intb){intc=a-b;returnc;}intmain(){inta,b;prin

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

doublefun(doublea,doubleb,intop){switch(op){case1:returna+b;break;case2:returna-b;break;case3:return

//fibonacci数列:1123581321...#include#includeintmain(void){longa=1;longb=1;intn;intk;printf("inputnumb

intHCF(intx,inty)//定义最大公约数函数{inti,change;if(x>y)//保证x是最小数{change=y;x=change;y=x;}for(i=x;i>=1;i--)if

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

第一题：#includevoidmain(){inta[10]={1,2,3,4,5,6,7,8,9,10},i,max,min;/*初始化的值任意定,只要是在整型范围内都行*/max=a[0];

#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min

占天时地利人和取九州四海财宝横批：财源不断

usingSystem;usingSystem.Collections.Generic;usingSystem.Linq;usingSystem.Text;namespaceConsoleApplic

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