下面分别是 n = 1， 2， 3 时的图案，给出 n 问你图案中有多少等边三角形，（三角形的三个点在顶点上）
画了一下午三角形，，，
以为是推公式，就手算前 6 项，结果答案越算越多，，，最后退出来前六项，盲猜公式。
ans[i] = n * (n+1) * (n+2) * (n+3) / 24
看别人题解是打表找的，下面是打表程序。（在坐标系中暴力找三点）
#include<iostream> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> //#include<stack> #include<queue> #include<deque> #include<set> #include<vector> #include<map> #include<functional> #define fst first #define sc second #define pb push_back #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 #define lc root<<1 #define rc root<<1|1 #define lowbit(x) ((x)&(-x)) #define mp make_pair using namespace std; typedef double db; typedef long double ldb; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PI; typedef pair<ll,ll> PLL; const db eps = 1e-6; const int mod = 1e9+7; const int maxn = 2e6+100; const int maxm = 2e6+100; const int inf = 0x3f3f3f3f; const db pi = acos(-1.0); vector<pair<double,double> >v; struct point{ double x, y; point(){} point(double a, double b){x = a; y = b;} }; double h = 0.5 *sqrt(3); double len(pair<double,double>a, pair<double,double>b){ return (a.fst-b.fst)*(a.fst-b.fst)+(a.sc-b.sc)*(a.sc-b.sc); } bool eq(double a, double b){ if(fabs(a-b)<1e-6)return true; return false; } int main(){ v.pb(mp(0,0)); for(int i = 1; i <= 2000; i++){ if(i&1){ for(int j = 0; j < i/2+1; j++){ v.pb(mp(j*1.0+0.5,-i*h)); v.pb(mp(-j*1.0-0.5,-i*h)); } } else{ v.pb(mp(0,-i*h)); for(int j = 1; j <= i/2; j++){ v.pb(mp(j*1.0,-i*h)); v.pb(mp(-j*1.0,-i*h)); } } int cnt = 0; for(int j = 0; j < (int)v.size(); j++){ for(int k = j+1; k < (int)v.size(); k++){ for(int g = k+1; g < (int)v.size(); g++){ if(eq(len(v[j],v[k]),len(v[k],v[g]))&&eq(len(v[j],v[k]),len(v[j],v[g])))cnt++; } } } printf("%d\n",cnt); } return 0; } /* 5 0 1 2 3 5 2 5 1 2 10 2 3 10 1 3 4 6 10 3 4 6 8 9 1 9 10 1 3 6 7 10 */