poj 1809 Regetni（數學）

Regetni

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 1595		Accepted: 615

Description

Background 
Hello Earthling. We're from the planet Regetni and need your help to make lots of money. Maybe we'll even give you some of it. 
You see, the problem is that in our world, everything is about integers. It's even enforced by law. No other numbers are allowed for anything. That said, it shouldn't surprise you that we use integer coordinate systems to plan our cities. So far only axis-aligned rectangular plots of land have been sold, but our professor Elgnairt recently had the revolutionary idea to sell triangular plots, too. We believe that the high society will love this concept and it'll make us rich. 
Unfortunately the professor patented his idea and thus we can't just do it. We need his permission and since he's a true scientist, he won't give it to us before we solve some damn riddle. Here's where you come in,because we heard that you're a genius. 

Problem 
The professor's riddle goes like this: Given some possible corners for the triangles, determine how many triangles with integral size can be built with them. Degenerated triangles with empty area (i.e. lines) have to be counted, too, since 0 is an integer. To be more precise, count the number of triangles which have as corners three different points from the input set of points. All points in a scenario will be distinct, i.e. there won't be duplicates. Here are some examples: 

Example a) shows a triangle with integral area (namely 3), b) shows one with non-integral size, c) shows a degenerated triangle with empty area (i.e. zero, so count it!), d) shows four points of which you can choose any three to build an integral area triangle and e) shows four points where you can't build any integral area triangles at all. 
Hint: The area A of a triangle with corners (x1, y1), (x2, y2) and (x3, y3) can be computed like this: 
A=|x1y2 - y1x2 + x2y3 - y2x3 + x3y1 - y3x1|/2 
Try to make clever use of this formula.

Input

The first line contains the number of scenarios. For each scenario, there is one line containing first the number N of distinct points in that scenario (0 <= N <= 10000) and after that N pairs of integers, each pair describing one point (xi, yi) with -100000 <= xi, yi <= 100000. All these numbers are separated by single blanks.

Output

Start the output for every scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the number of triangles with integral area whose three distinct corners are among the points given. Terminate the output for each scenario with a blank line.

Sample Input

6
3 0 0 2 0 1 -3
3 0 0 2 1 1 -3
3 0 0 2 2 3 3
4 0 0 2 0 0 2 2 2
4 0 0 1 0 0 1 1 1
9 0 0 0 1 0 2 1 0 1 1 1 2 2 0 2 1 2 2

Sample Output

Scenario #1:
1

Scenario #2:
0

Scenario #3:
1

Scenario #4:
4

Scenario #5:
0

Scenario #6:
48

Source

TUD Programming Contest 2003, Darmstadt, Germany

題意：給定一些整數座標的點，問能組成的整數面積的情況數

題解：最大限度的利用題目所給的公式，分析公式可知，組合數只和點的奇偶有關，那麼只需記錄4種情況的點的數目，奇奇，奇偶，偶奇，偶偶，然後把所有無順序組合情況列出來，排除不能得出整數結果的，然後直接打公式出來就行了

#include<stdio.h>
long long x,y,ans,zz,zo,oz,oo;
int main()
{
    int i,n,t,cas=1;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        ans=zz=zo=oz=oo=0;
        for(i=0;i<n;i++)
        {
            scanf("%I64d%I64d",&x,&y);
            if(x&1)
            {
                if(y&1) oo++;
                else oz++;
            }
            else
            {
                if(y&1) zo++;
                else zz++;
            }
        }
        ans+=zz*(zz-1)*(zz-2)/6;
        ans+=zo*(zo-1)*(zo-2)/6;
        ans+=oz*(oz-1)*(oz-2)/6;
        ans+=oo*(oo-1)*(oo-2)/6;

        ans+=zz*(zz-1)*zo/2;
        ans+=zz*(zz-1)*oz/2;
        ans+=zz*(zz-1)*oo/2;

        ans+=zo*(zo-1)*zz/2;
        ans+=zo*(zo-1)*oz/2;
        ans+=zo*(zo-1)*oo/2;

        ans+=oz*(oz-1)*zz/2;
        ans+=oz*(oz-1)*zo/2;
        ans+=oz*(oz-1)*oo/2;

        ans+=oo*(oo-1)*zz/2;
        ans+=oo*(oo-1)*zo/2;
        ans+=oo*(oo-1)*oz/2;

        printf("Scenario #%d:\n",cas++);
        printf("%I64d\n\n",ans);
    }

    return 0;
}