bzoj 3505 [Cqoi2014]数三角形——排列组合

题目：https://www.lydsy.com/JudgeOnline/problem.php?id=3505

好题！一定要经常回顾！

那个 一条斜线上的点的个数是其两端点横坐标之差和纵坐标之差的gcd-1 真是很妙。

https://blog.csdn.net/u012288458/article/details/48624859

https://www.cnblogs.com/Var123/p/5377616.html

然而还可以递推？https://www.cnblogs.com/liu-runda/p/5993244.html 反正没管……  

#include<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
int n,m;
ll ans;
ll C(int x)
{
  return (ll)x*(x-1)*(x-2)/6;
}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main()
{
  scanf("%d%d",&n,&m);n++;m++;
  ans=C(n*m)-n*C(m)-m*C(n);
  for(int i=1;i<n;i++)// bh node from 0 to n-1
    for(int j=1;j<m;j++)
      {
    ll num=gcd(i,j)-1;
    if(num>=1)ans-=num*(n-i)*(m-j)*2;
      }
  printf("%lld",ans);
  return 0;
}