题目链接：
题意：求n * m矩形内，最多能组成几个三角形
这题和UVA 1393类似，把总情况扣去三点共线情况，那么问题转化为求三点共线的情况，对于两点，求他们的gcd - 1，得到的就是他们之间有多少个点，那么情况数就能够求了，然后还是利用容斥原理去计数，然后累加出答案
代码：
#include#include #include using namespace std;const int N = 1005;long long n, m, dp[N][N];int cas;long long gcd(long long a, long long b) { if (b == 0) return a; return gcd(b, a % b);}void init() { cas = 0; for (int i = 2; i <= 1000; i++) for (int j = 2; j <= 1000; j++) dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + gcd(i, j) - 1; for (int i = 2; i <= 1000; i++) for (int j = 2; j <= 1000; j++) dp[i][j] += dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];}long long C(long long n, long long m) { if (m > n) return 0; m = min(m, n - m); long long ans = 1; for (long long i = 0; i < m; i++) ans = ans * (n - i) / (i + 1); return ans;}int main() { init(); while (~scanf("%lld%lld", &n, &m) && n || m) { n++; m++; printf("Case %d: %lld\n", ++cas, C(n * m, 3) - n * C(m, 3) - m * C(n, 3) - dp[n - 1][m - 1] * 2); } return 0;}