[highlight]Theory[/highlight]
Picks formulae :::
ans=area of polygon-B/2+1;
Given a polygon whose vertices are of integer coordinates (lattice points), count the number of lattice points within that polygon.
Pick’s theoreom states that
where;
is the area of the polygon.
is the number of lattice points on the exterior
Area of a polygon:
First, number the vertices in order, going either clockwise or counter-clockwise, starting at any vertex.
The area is then given by the formula
[highlight]Solution:[/highlight]
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#include <iostream> #include <stdio.h> #include <vector> using namespace std; struct point{ long long x; long long y; }; vector<point> vb; long long gcd(long long dx,long long dy) { if(dx==0) return dy; return gcd(dy%dx,dx); } int main() { int n; long long a,b; double area; long long B; point p; //cout<<gcd(45,30)<<"\n"; //cout<<gcd(216,48); while(cin>>n) { if(n==0) break; vb.clear(); B=0; area=0; for(int i=0;i<n;i++) { cin>>a>>b; p.x=a; p.y=b; vb.push_back(p); /*if(i>=1) { area+=(vb[i-1].x*vb[i].y+vb[i].x*vb[i-1].y); long long dx=vb[i-1].x-vb[i].x; long long dy=vb[i-1].y-vb[i].y; if(dx<0) dx*=-1; if(dy<0) dy*=-1; B+=gcd(dx,dy); }*/ } for(int i=1;i<=n;i++) { area+=(vb[i-1].x*vb[i%n].y-vb[i%n].x*vb[i-1].y); long long dx=vb[i-1].x-vb[i%n].x; long long dy=vb[i-1].y-vb[i%n].y; if(dx<0) dx*=-1; if(dy<0) dy*=-1; B+=gcd(dx,dy); } //area+=(vb[n-1].x*vb[0].y+vb[0].x*vb[n-1].y); area/=2; if(area<0) area*=-1; //long long dx=vb[n-1].x-vb[0].x; //long long dy=vb[n-1].y-vb[0].y; /*if(dx<0) dx*=-1; if(dy<0) dy*=-1; B+=gcd(dx,dy); */ //pick's formulae ans=area of polygon-B/2+1 long long ans= (long long )(area-B/2+1); cout<<ans<<"\n"; } return 0; } |