第2个回答 2013-04-08
void main()
{
int a,b,c,p,s
p=(a+b+c)/2
s= S=√[p(p-a)(p-b)(p-c)]
scanf("a""b""c")
print (“s"/n)
}
第3个回答 2013-04-08
#include《stdio.h》
{
int a,b,c;//三边长
float C,d,e,f;
float s;//面积
printf("请输入三边长\n");
scanf("%d %d %d",&a,&b,&c);
C=1/2(a+b+c);
d=C-a;
e=C-b;
f=C-c;
s=sqrt(C*d*e*f);
printf("三角形的面积为:%f",s);
}
第4个回答 2013-04-08
#include "stdio.h"#include "Math.h"
double Mianji(double a,double b,double c)
{
double p;
p=(a+b+c)/2;
return sqrt(p*(p-a)*(p-b)*(p-c));
}
boolean isSanJiaoXing(double a,double b,double c){
if((a+b>c)&&(a+c>b)&&(b+c>a))
return 1;
return 0;
}
void main()
{
double a,b,c,p;
double mianji;
printf("plz input a:\n");
scanf(%f%,&a);
printf("plz input b:\n");
scanf(%f%,&b);
printf("plz input c:\n");
scanf(%f%,&c);
if(isSanJiaoXing(a,b,c))
{
mianji=Mianji(a,b,c);
printf("mian ji is:%.3f\n",mianji);
}
else
printf("shuru bu hefa\n");
}