Question

C++输入3个整数,输出它们的平均值,保留3位小数.

Answer 1

楼上写的麻烦了
#include <iostream>
using namespace std;
int main(){
int a,b,c;

cin >> a >> b >> c;

double average=(a+b+c)/3.0; //是3.0 不能是3

cout.precision(3); //控制输出位数为3

cout << fixed << average << endl; //输出位数需要fixed控制
return 0;
}

Answer 2

#include <iostream>
#include <iomanip>
using namespace std;
double average(double a, double b, double c)
{
double result=(a+b+c)/3.000;
return(result);
}

int main()
{
int a, b, c;
cin>>a>>b>>c;
cout<<setiosflags(ios::fixed)<<setprecision(3)<<average(a, b, c)<<endl;
return(0);
}

Answer 3

//
cout.precision
void
ccoutformatsample::p_17_6(void)
{
cout
<<
_t("
17.6
设置浮点数的显示精度")
<<
endl
<<
endl;
float
f1
=
23.3232;
float
f2
=
1.9
+
8.0
/
9.0;
cout
<<
"f1
=
"
<<
f1
<<
endl;
cout
<<
"f2
=
"
<<
f2
<<
endl;
//
设置精度为2
cout.precision(2);
cout
<<
"f1
=
"
<<
f1
<<
endl;
cout
<<
"f2
=
"
<<
f2
<<
endl;
//
解除设定
cout.precision();
cout
<<
endl
<<
endl;
}
这段资料对你应该有用吧。
以上搜自http://blog.csdn.net/revision/archive/2008/01/10/2032797.aspx