Pick公式的應(yīng)用。先介紹一下Pick公式：
a = e / 2 + i - 1
a為多邊形（頂點(diǎn)都在格點(diǎn)上）面積，e為多邊形邊上的格點(diǎn)數(shù)，i為多邊形內(nèi)部的格點(diǎn)數(shù)。
題目給出三點(diǎn)坐標(biāo)，邊上的格點(diǎn)數(shù)可用gcd求得，剩下的事就是解方程了。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-9-18 21:19:33
File Name: pku2954.cpp
Description:
************************************************************************/
#include <iostream>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
int main()
{
int ax, ay, bx, by, cx, cy;
while (scanf("%d%d%d%d%d%d", &ax, &ay, &bx, &by, &cx, &cy), !(ax == 0 && ay == 0 && bx == 0 && by == 0 && cx == 0 && cy == 0))
{
int tax = bx - cx, tay = by - cy;
int tbx = cx - ax, tby = cy - ay;
int tcx = ax - bx, tcy = ay - by;
int b = gcd(abs(tcx), abs(tcy)) + gcd(abs(tax), abs(tay)) + gcd(abs(tbx), abs(tby));
int a = abs(tax * tby - tay * tbx);
printf("%d\n", (a + 2 - b) / 2);
}
return 0;
}