根据输入的三个边长值(整型值),判断能否构成三角形.cyvyan\

voidswap(int*x,int*y){inttmp=*x;*x=*y;*y=tmp;}

#includedoublearea(doublea,doubleb,doubleh){return0.5*(a+b)*h;//二分之一上底加下底的和乘以高}intmain(void){doublea

y+=j*1.0/(i*i);

 #include<stdio.h>void main(){     int x,y; print

a=Val(inputbox("输入a"))b=Val(inputbox("输入b"))c=Val(inputbox("输入c"))Ifa+b>cAndb+c>aAndc+a>bThen'能构成Ifa

#includemain(){intx,y,m,n;scanf("%d,%d",&x,&y);if(x>y,m=x-y)printf("%d\n",m);else(n=y-x)printf("%d\n

#includeintmain(void){intn=20,i,x,sum=0,a=0;for(i=1;i0)sum+=x,a++;}printf("%d\n",sum/a);return0;}楼上的

voidfun(inta,intb,intc)上面这句改成intfun(inta,intb,intc)//elseretrun1这句LZ如果不是抄错的话应该改成elsereturn1;

首先要看用什么平台vb和c语言是不一样的,还有其他的我写个c语言的吧main(){inta,b,c,d,e,sum;a=10;b=20;c=12;sum=a+b+c;d=a*b*c;e=(a+b+c)

#includeintmain(void){intflag=1,i,n,an;floatresult=0;printf("pleaseinputn:\n");scanf("%n",&n);for(i=

#includeintmax(inti,intj){return(i>j)?i:j;}intmin(inti,intj){return(i

voidfun(int*a,int*b){intt;t=*a;*a=*b;*b=t;}你这是作业题吧!

//看好记事本写的Judge.javapublicclassJudge{publicstaticvoidmain(String[]args){try{if(args.length!=3){Sy

#includeusingnamespacestd;intmain(){inta,b,c,temp;couta>>b>>c;if(a>b){temp=a;a=b;b=temp;}

#include"stdio.h"doublesum(intm){\x09inti;\x09doublet;\x09for(i=2,t=1.0;i

PublicFunctionArea(aAsDouble,bAsDouble,cAsDouble)AsDoubleIfa+bMSGBOX"要求任意两边之和大于第三边"ExitFunctionendif

#include#includevoidmain(){floata,b,c,s,p;printf("pleaseinputthea,b,c:");scanf("%f%f%f",&a,&b,&c);p=

你用到了sqrt()函数了,这个函数包含在math.h头文件里面,你在最前面加一句：#include就可以了

#includeusingnamespacestd;intmain(){inta,b;cin>>a>>b;intl;l=2*(a+b);cout再问：看不懂，能不能简单一点的再答：#includeus

intfun(intm){inty=1;for(inti=2;i