0
0
已采纳
for(int i=0;i<=n-1;i++)
{
d[i]=ceil(sqrt(a[i]*a[i]+b[i]*b[i])/5*2+c[i]*10+c[i]*5);
sum+=d[i];
}
0
0
for(int i=1;i<=n;i++)
{
cin>>x>>y>>rs;
k=sqrt(x*x+y*y)/5+rs*10+sqrt(x*x+y*y)/5+rs*5;
sum+=k;
}
printf("%d",int(sum+0.9));
0
0
0
0
