bzoj3505: [Cqoi2014]数三角形

首先共有(n+1)*(m+1)个点，所以先++n,++m

不考虑三点共线的情况，有(n*m)*(n*m-1)*(n*m-2)/6个

三点都在水平和竖直的有m*(m-1)*(m-2)*n/6+n*(n-1)*(n-2)*m/6个

然后考虑斜着的

枚举每个点作为第一个点，使后两个点坐标大于第一个

过第一个点做个水平和竖直的线，右上和左下的矩形可用

可用dp求出一个矩阵中第一个点在角上，三点共线的个数

#include <cstdio>
#include <algorithm>
using namespace std;
#define MAXN 1003
#define MAXP 170
int gcd(int x, int y)
{
while (y)
{
int t = x % y;
x = y;
y = t;
}
return x;
}
int n, m;
int a[MAXN][MAXN], ct[MAXN][MAXN];
long long f[MAXN][MAXN];
int main()
{
scanf("%d%d", &n, &m);
++n, ++m;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j)
{
int g = gcd(i, j);
int x = i / g, y = j / g;
a[i][j] = ct[x][y]++;
}
for (int i = 1; i <= n; ++i)
f[i][1] = f[i - 1][1] + a[i][1];
for (int i = 1; i <= m; ++i)
f[1][i] = f[1][i - 1] + a[1][i];
for (int i = 2; i <= n; ++i)
for (int j = 2; j <= m; ++j)
f[i][j] = f[i - 1][j] + f[i][j - 1] - f[i - 1][j - 1] + a[i][j];
int N = n * m;
long long ans = (long long)N * (N - 1) * (N - 2) / 6;
ans -= (long long)m * (m - 1) * (m - 2) / 6 * n + (long long)n * (n - 1) * (n - 2) / 6 * m;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j)
ans -= f[n - i][j - 1] + f[n - i][m - j];
printf("%lld\n", ans);
return 0;
}