UVALive - 3295

Posted 2020-10-01

给定n行m列个正方形，求可以组成多少个三角形。

如1×2，有18个。

总的点有N个，那个答案就是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。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <string>
 6 #include <queue>
 7 #include <vector>
 8 #include <set>
 9 #include <map>
10 #define ll long long
11 #define INF 0x3f3f3f3f
12 #define lowvit(x) x&(-x)
13 #define N 1010
14 #define M 110
15 using namespace std;
16 ll gcd(ll a, ll b) {
17     return b ? gcd(b, a%b): a;
18 }
19 int main() {
20     ios::sync_with_stdio(false);
21     ll a, b;
22     int k = 1;
23     while(cin>>a>>b) {
24         if(a==0&&b==0)break;
25         ll n = (a+1)*(b+1);
26         ll sum1 = n*(n-1)*(n-2)/6;
27         ll sum2 = (a+1)*(b+1)*(b)*(b-1)/6 + (b+1)*(a+1)*a*(a-1)/6;
28         ll sum3 = 0;
29         for(int i = 2; i <= a; i ++) {
30             for(int j = 2; j <= b; j ++) {
31                 sum3 += (gcd(i,j)-1)*(a-i+1)*(b-j+1);
32             }
33         }
34         ll ans = sum1 - sum2 - 2*sum3;
35         printf("Case %d: %lld\\n",k++,ans);
36     }
37     return 0;
38 }