公式很容易猜出，見代碼
#include <iostream>
int gcd(int a, int b) 
{
if (b == 0) return a;
else return gcd(b, a % b);
}
int main() {
int x[110], y[110], dx[110], dy[110], f[110];
int s, k, kase, n, e, i, j;
scanf("%d", &kase);
for (k = 1; k <= kase; k++)
{
scanf("%d", &n);
memset(f, 0, sizeof(f));
x[0] = 0; y[0] = 0; s = 0; e = 0;
for (i = 1; i <= n; i++)
{
scanf("%d%d", &dx[i], &dy[i]);
x[i] = x[i - 1] + dx[i];
y[i] = y[i - 1] + dy[i];
e += gcd(abs(dx[i]), abs(dy[i]));
s += x[i - 1] * y[i] - x[i] * y[i - 1];
}
j = s + 2 - e;
printf("Scenario #%d:\n", k);
printf("%d %d %.1lf\n\n", j / 2, e ,(double) s / 2);
}
return 0;
}