poj1265皮克定理

給出每次所走的距離dx,dy;   求多邊形內部點的個數、邊上點的個數、面積.

 

#include<iostream>#include<cmath>#include<iomanip>using namespace std;struct point{   int x,y;};int gcd(int m,int n){if(n==0) return m;return gcd(n,m%n);}int bian(point A,point B)//算出點A和點B組成的線段上的點{return gcd(abs(A.x-B.x),abs(A.y-B.y));}int f(point a[],int n)//求n邊形的面積{int i,ans=0;a[n]=a[0];for(i=0;i<n;i++)ans+=(a[i].x*a[i+1].y-a[i].y*a[i+1].x);if(ans<0) ans=-ans;return ans;}int main(){  int s,ans,t,n,i,j=0;   point a[105];   cin>>t;   while(j<t)   { cin>>n;     for(i=0;i<n;i++)      {cin>>a[i].x>>a[i].y;       if(i>0) {a[i].x+=a[i-1].x; a[i].y+=a[i-1].y;}      }      for(s=i=0;i<n;i++)      { if(i==n-1) s+=bian(a[i],a[0]);      else      s+=bian(a[i],a[i+1]);      }            ans=f(a,n);      cout<<"Scenario #"<<++j<<":"<<endl<<(ans-s)/2+1<<" "<<s<<" "<<fixed<<setprecision(1)<<ans/2.0<<endl<<endl;   }return 0;}