lightoj1418Trees on My Island【pick公式】

Q - Trees on My Island
Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit Status Practice LightOJ
1418

Description

I have bought an island where I want to plant trees in rows and columns. So, the trees will form a rectangular grid and each of them can be thought of having integer coordinates by taking a suitable grid point as the origin.

But, the problem is that the island itself is not rectangular. So, I have identified a simple polygonal area inside the island with vertices on the grid points and have decided to plant trees on grid points lying strictly inside the polygon.



Figure: A sample of my island

For example, in the above figure, the green circles form the polygon, and the blue circles show the position of the trees.

Now, I seek your help for calculating the number of trees that can be planted on my island.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer N (3 ≤ N ≤ 10000) denoting the number of vertices of the polygon. Each of the next N lines contains two integers xi yi(-106 ≤ xi,
yi ≤ 106) denoting the co-ordinate of a vertex. The vertices will be given in clockwise or anti-clockwise order. And they will form a simple polygon.

Output

For each case, print the case number and the total number of trees that can be planted inside the polygon.

Sample Input

1

9

1 2

2 1

4 1

4 3

6 2

6 4

4 5

1 5

2 3

Sample Output

Case 1: 8
题意：求多边形内的点数
解题思路：pick公式

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct point{
long long x,y;
}A[10010];
long long gcd(long long a,long long b){
return b==0?a:gcd(b,a%b);
}
int main()
{
long long t,k=1,n,i,j;
scanf("%lld",&t);
while(t--){
scanf("%lld",&n);
for(i=0;i<n;++i){
scanf("%lld%lld",&A[i].x,&A[i].y);
}
long long dot=0,S=0;
for(i=0;i<n;++i){
dot+=gcd(abs(A[(i+1)%n].x-A[i].x),abs(A[(i+1)%n].y-A[i].y));
S+=A[(i+1)%n].y*A[i].x-A[i].y*A[(i+1)%n].x;
}
printf("Case %lld: %lld\n",k++,(abs(S)+2-dot)/2);
}
return 0;
}