hdu5563 Clarke and five-pointed star

来源：互联网发布：工程模式设置电信网络编辑：程序博客网时间：2024/05/16 05:21

思路：正五边形的判断。在正五边形中任意两点之间的距离只有两种情况，所以求出所有点之间的距离排序，前后比较纪录有多少个不同距离的边长。

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>// #define DEBUG#ifdef DEBUG#define debug(...) printf( __VA_ARGS__ )#else#define debug(...)#endif#define MEM(x,y) memset(x, y,sizeof x)using namespace std;typedef long long LL;typedef unsigned long long ULL;typedef pair<double,double> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;ii p[6];const double eps = 1e-6;double mul(double x){return x*x;}double GetDis(const ii& A,const ii& B){return sqrt(mul(A.first - B.first) + mul(A.second - B.second));}double dis[12];int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int t;scanf("%d",&t);while(t--){int cnt = 1;for (int i = 1;i <= 5;++i)scanf("%lf%lf",&p[i].first,&p[i].second);for (int i = 1;i <= 5;++i){for (int j = i + 1;j <= 5;++j)dis[cnt++] = GetDis(p[i], p[j]);}sort(dis + 1,dis + cnt);int ans = 1;for (int i = 2;i < cnt;++i){if (fabs(dis[i] - dis[i-1]) > eps) ans++;}if (ans <= 2) puts("Yes");else puts("No");}return 0;}

0 0