Area POJ - 1265
题意: 给定一个多边形的相邻点的dx,dy. 求在多边形内部点的个数in,多边形边界上的个数on,多边形的面积s
思路: s代表多边形面积 , on=abs(gcd(dx,dy)), in=(2s+2-on)/2; 求面积的时候/2 *2抵消
对于S求叉击,用long long 保存, 减少精度误差
#include<cstdio>
#include<vector>
#include<cmath>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
#define PI acos(-1.0)
#define pb push_back
#define F first
#define S second
#define x adsfadwsfasdfas
#define sx dsafasfaseer
#define sy dsafsfaseer
#define sx dsafasfaser
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=1e6+6;
const int MOD=1e9+7;
template <class T>
bool sf(T &ret){ //Faster Input
char c; int sgn; T bit=0.1;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
if(c==' '||c=='\n'){ ret*=sgn; return 1; }
while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
ret*=sgn;
return 1;
}
struct Point{
ll x,y;
Point(ll x=0.0, ll y=0.0) : x(x), y(y) {}
Point operator-(const Point &rhs)const{
return Point(x-rhs.x,y-rhs.y);
}
};
typedef Point Vector;
ll cross(Vector A,Vector B){
return A.x*B.y-A.y*B.x;
}
char s[N];
int main(void){
int T;
sf(T);
int ks=0;
while(T--){
int n;
sf(n);
ll sx=0,sy=0;
long long ans=0;
ll on=0,in=0;
for(int i=1;i<=n;i++){
int x,y;
sf(x),sf(y);
ans+= cross({sx,sy},{sx+x,sy+y});
sx+=x,sy+=y;
on+=abs(__gcd(x,y));
}
in=(ans+2-on)/2;
printf("Scenario #%d:\n",++ks);
printf("%lld %lld %lld",in,on,ans/2);
if(ans&1) printf(".5\n\n");
else printf(".0\n\n");
}
return 0;
}