UVALive 3295 Counting Triangles

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1296

题意：给出一个a*b的网格，在网格上取不共线的三点构成三角形，求三角形总数。
分析：就是一一道简单的组合数计算题目，设总结点数为n，则取三个节点的个数为C（n，3），

然后减去横向、竖向、斜向的三点共线的个数即可，斜线三点共线等价于所枚举的矩形的长宽成倍数关系，即gcd不为1

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#include <queue>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
bool cmp(int x,int y)
{
    return x>y;
}
const int N=10005;
const int mod=1e9+7;
int main(){
    ll a, b;
    int cas = 1;
    while(scanf("%lld%lld", &a, &b)!=EOF && (a||b)){
        ll n = (a+1)*(b+1);
        ll sum1 = n*(n-1)*(n-2)/6;        //C(n,3)
        ll sum2 = (b+1)*(a+1)*a*(a-1)/6 + (a+1)*(b+1)*b*(b-1)/6; //横向或竖向三点共线的个数
        ll sum3 = 0;        //斜线上三点共线的个数的一半
        int i, j;
        for(i=2; i<=a; i++)
            for(j=2; j<=b; j++)
                sum3 += (gcd(i,j)-1) * (a-i+1) * (b-j+1);
        ll ans = sum1 - 2*sum3 - sum2;
        printf("Case %d: %lld
", cas++, ans);
    }
    return 0;
}