CodeForces 78D Archer's Shot

二分。

统计过程如下图：

先统计红线上的个数，然后统计绿线上的个数，然后统计咖啡色线上的个数......一个一个往下统计就可以了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
}

double r;
double sqrt3=sqrt(3.0);

double dis(double x,double y) { return x*x+y*y; }

bool check(double x,double y)
{
    if(dis(x,y+1)-eps>r*r) return 0;
    if(dis(x,y-1)-eps>r*r) return 0;
    if(dis(x+sqrt3/2,y+0.5)-eps>r*r) return 0;
    if(dis(x+sqrt3/2,y-0.5)-eps>r*r) return 0;
    if(dis(x-sqrt3/2,y+0.5)-eps>r*r) return 0;
    if(dis(x-sqrt3/2,y-0.5)-eps>r*r) return 0;
    return 1;
}

int main()
{
    while(~scanf("%lf",&r))
    {
        LL ans=0, L=1,R=(LL)2000000,pos;
        while(L<=R)
        {
            LL mid=(L+R)/2;
            if(check((mid-1)*sqrt3,0)) pos=mid, L=mid+1;
            else R=mid-1;
        }
        LL cnt=0; double x=0,y=-3;
        while(1)
        {
            if(!check(x,y)) break;
            LL L=1,R=(LL)2000000,pp;
            while(L<=R)
            {
                LL mid=(L+R)/2;
                if(check(x-(mid-1)*sqrt3/2,y-(mid-1)*1.5)) pp=mid, L=mid+1;
                else R=mid-1;
            }
            pp=1+(pp-1)*2, cnt=cnt+pp, y=y-3;
        }
        ans=6*(pos-1)+1+6*cnt;
        printf("%lld\n",ans);
    }
    return 0;
}