c语言编写一个计算两个输入正整数n1 与n2 的最大公约数和最小公倍数的程序
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#include"stdio.h"intmain(){\x09inti,j,n;\x09inta[12];\x09intmin,mx;\x09scanf("%d",&n);\x09for(i=0;i
#includemain(){floata,b,c;scanf("%f%f",&a,&b);c=a+b;printf("%.2f\n",c);}
#includevoidmain(){inta,b;printf("Inputtwointegers:");scanf("%d%d",&a,&b);printf("和:a+b=%d\n",a+b);p
#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for
#include"stdio.h"main(){intm,n,t,h,a,b,q;printf("输入m,n的值:");scanf("%d%d",&m,&n);a=m;b=n;if(n>m){t=m;
首先明确gbs(最小公倍数)=m*n/gys(最大公约数);然后求最大公约数用欧几里得辗转相除法;代码很短的.#include"stdio.h"intgys(intm,intn){returnn==0
//别说100位,1000位都没问题,给你个例子.#include#definePRECISION2800#defineFRACTION1000#defineGROUP4#defineINITIALV
#include"stdio.h"intmain(){inta[10][10]={0};intn,i,j,sum1=0,sum2=0;scanf("%d",&n);for(i=0;i
希望有用,敬请采纳^_^#include#includeintmain(){voidroot2(doublea,doubleb,doubledisc);//定义方程有两个根时的函数voidroot1(
#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}
#includeintgongyue(intm,intn){intr;if(m==n)returnm;elsewhile((r=m%n)!=0){m=n;n=r;}returnn;}voidmain(
给你写了个,运行通过,你看看吧,记得采纳哦O(∩_∩)O~#includeintmain(){\x09intnum,i=0;\x09printf("pleaseinputanumble:");\x09
输入两个整数,中间空格:#include<stdio.h>main(){\x09int a,b;\x09printf("Input a,b:\n")
#include"stdio.h"//voidmain(void){inta,b,c;printf("请输入两个十进制整数!\na=");scanf("%d",&a);printf("b=")
#includeintgetArea(intlen){\x09returnlen*len;}intmain(){\x09intn;\x09intarea;\x09\x09printf("请输入边长:\
#include#include#defineE2.7182voidmain(void){floatx;floaty;scanf("%d",&x);if(x>=-1)y=pow(x,2)+1;
#include <stdio.h>int main(){ float x, tax = 0;
#includevoidmain(){\x09intm,n,k,min,max,i;\x09scanf("%d%d",&m,&n);\x09if(n0;i++)\x09{\x09\x09if((m%i
#include <stdio.h>main(){ double grade[200]; double sum,avg,temp;&nbs
#includeintmain(){intlength,width,perimeter;printf("长:");scanf("%d",&length);printf("宽:");scanf("%d"