关于floor函数
原题链接：71101.Force
2025-09-07 19:31:31
发布于：浙江
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为什么我在考试时 floor 函数没用 int 强制转换就爆零了，今天加了个int就A了，虽然返回值是double类型，但输出应该是一致的啊，难不成数据全都大到用e表示了吗？
0pts:
#include<bits/stdc++.h>
#define int long long
#define ldouble long double
using namespace std;
const double pi=acos(-1.0);
int n;
double a=0.0,b=0.0;
/*ldouble sq(ldouble x,ldouble y)
{
return sqrt(x*x+y*y);
}*/
signed main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
double x,f,r;
cin>>x>>f;
r=x*pi/180.0;
a+=f*cos(r);
b+=f*sin(r);
}
double R=hypot(a,b);
double ans;
if(R<1e-12)ans=0.0;
ans=(fabsl(R)<1e-12?0:atan2(b,a)*180.0/pi);
while(ans<0)ans+=360.0;
cout<<floor(R+1e-12)<<" "<<floor(ans+1e-12)<<endl;
return 0;
}
AC:
#include<bits/stdc++.h>
#define int long long
#define ldouble long double
using namespace std;
const double pi=acos(-1.0);
int n;
double a=0.0,b=0.0;
/*ldouble sq(ldouble x,ldouble y)
{
return sqrt(x*x+y*y);
}*/
signed main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
double x,f,r;
cin>>x>>f;
r=x*pi/180.0;
a+=f*cos(r);
b+=f*sin(r);
}
double R=hypot(a,b);
double ans;
if(R<1e-12)ans=0.0;
ans=(fabsl(R)<1e-12?0:atan2(b,a)*180.0/pi);
while(ans<0)ans+=360.0;
cout<<(int)floor(R+1e-12)<<" "<<(int)floor(ans+1e-12)<<endl;
return 0;
}
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