题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=4493
一个double 如果就用cout 貌似会用科学技术法;
#include<iostream>
using namespace std;
int x1(double k)
{
k*=10;
return (int)k%10;
}
int x2(double k)
{
k*=100;
return (int)k%10;
}
int x3(double k)
{
k*=1000;
return (int)k%10;
}
int main()
{
int T;
cin>>T;
while(T--)
{
double ans=0;
for(int i=0;i<12;i++)
{
double temp;
cin>>temp;
ans+=temp;
}
ans/=12;
if(x3(ans)>=5) ans+=0.01;
cout<<"$"<<(long long )ans;
int a1=x1(ans);
int a2=x2(ans);
if(a1==0&&a2==0) cout<<endl;
else if(a1!=0&&a2==0) cout<<"."<<a1<<endl;
else cout<<"."<<a1<<a2<<endl;
// cout<<"$"<<ans<<endl;
//这样就会错
}
}