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=;
 const int mod=1e9+;
 int main(){
     ll a, b;
     int cas = ;
     while(scanf("%lld%lld", &a, &b)!=EOF && (a||b)){
         ll n = (a+)*(b+);
         ll sum1 = n*(n-)*(n-)/;        //C(n,3)
         ll sum2 = (b+)*(a+)*a*(a-)/ + (a+)*(b+)*b*(b-)/; //横向或竖向三点共线的个数
         ll sum3 = ;        //斜线上三点共线的个数的一半
         int i, j;
         for(i=; i<=a; i++)
             for(j=; j<=b; j++)
                 sum3 += (gcd(i,j)-) * (a-i+) * (b-j+);
         ll ans = sum1 - *sum3 - sum2;
         printf("Case %d: %lld\n", cas++, ans);
     }
     return ;
 }