HDU 6055 - Regular polygon
阿新 • • 發佈:2017-07-27
clas name n) cnblogs 多少 its light ++ space
/* HDU 6055 - Regular polygon [ 分析,枚舉 ] 題意: 給出 x,y 都在 [-100, +100] 範圍內的 N 個整點,問組成的正多邊形的數目是多少 N <= 500 分析: 分析可知,整點組成的正多邊形只能是正方形 故枚舉兩個點,驗證剩下兩個點的位置 坑點: 由於點的範圍是 [-100, +100],故經過計算得出的點的範圍可能是 [-300,+300],註意越界 編碼時長:46分鐘(-1) */ #include <bits/stdc++.h> using namespace std; int n; bool mp[1005][1005]; int ans; int x[505], y[505]; void solve(int x1, int y1, int x2, int y2) { if (x1 > x2) swap(x1, x2), swap(y1, y2); int x3, y3, x4, y4; x3 = x1 - (y2-y1); y3 = y1 + x2-x1; x4 = x2 - (y2-y1); y4 = y2 + x2-x1; if (mp[x3][y3] && mp[x4][y4]) ans++; x3 = x1 + y2-y1; y3 = y1 - (x2-x1); x4 = x2 + y2-y1; y4 = y2 - (x2-x1); if (mp[x3][y3] && mp[x4][y4]) ans++; } int main() { while (~scanf("%d", &n)) { memset(mp, 0, sizeof(mp)); for (int i = 1; i <= n; i++) { scanf("%d%d", &x[i], &y[i]); x[i] += 500, y[i] += 500; mp[x[i]][y[i]] = 1; } ans = 0; for (int i = 1; i <= n; i++) for (int j = i+1; j <= n; j++) solve(x[i], y[i], x[j], y[j]); printf("%d\n", ans/4); } }
HDU 6055 - Regular polygon