Kirai[HDOJ5365]Run
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5365
众所周知，整点是围不成正三角形正五边形正六边形的，所以我们只需暴力出它可以围成几个正四边形。
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <cctype> 8 #include <queue> 9 #include <map> 10 #include <set> 11 #include <stack> 12 #include <list> 13 #include <vector> 14 15 using namespace std; 16 17 typedef struct Node { 18 int x; 19 int y; 20 }; 21 Node po[22]; 22 int edge[11]; 23 int n; 24 25 int dis(int x, int y) { 26 return x * x + y * y; 27 } 28 int solve() { 29 memset(edge, 0, sizeof(edge)); 30 int ans = 0; 31 for(int i = 0; i < n; i++) { 32 for(int j = i+1; j < n; j++) { 33 for(int p = j+1; p < n; p++) { 34 for(int q = p+1; q < n; q++) { 35 edge[0] = dis(po[i].x-po[j].x, po[i].y-po[j].y); 36 edge[1] = dis(po[i].x-po[p].x, po[i].y-po[p].y); 37 edge[2] = dis(po[i].x-po[q].x, po[i].y-po[q].y); 38 edge[3] = dis(po[j].x-po[p].x, po[j].y-po[p].y); 39 edge[4] = dis(po[j].x-po[q].x, po[j].y-po[q].y); 40 edge[5] = dis(po[p].x-po[q].x, po[p].y-po[q].y); 41 sort(edge, edge+6); 42 if(edge[0] == edge[1] && edge[1] == edge[2] && edge[2] == edge[3] && edge[4] == edge[5]) { 43 ans++; 44 } 45 } 46 } 47 } 48 } 49 return ans; 50 } 51 int main() { 52 // freopen("in", "r", stdin); 53 while(~scanf("%d", &n)) { 54 for(int i = 0; i < n; i++) { 55 scanf("%d %d", &po[i].x, &po[i].y); 56 } 57 printf("%d\n", solve()); 58 } 59 }
