本文主要是介绍poj 1675 Happy Birthday!，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
与原点连线的最大夹角不小于120即可。
/** Author: stormdpzh* Created Time: 2012/7/19 16:01:35* File Name: poj_1675.cpp*/
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <functional>#define sz(v) ((int)(v).size())
#define rep(i, n) for(int i = 0; i < n; i++)
#define repf(i, a, b) for(int i = a; i <= b; i++)
#define repd(i, a, b) for(int i = a; i >= b; i--)
#define out(n) printf("%d\n", n)
#define mset(a, b) memset(a, b, sizeof(a))
#define wh(n) while(1 == scanf("%d", &n))
#define whz(n) while(1 == scanf("%d", &n) && n != 0)
#define lint long longusing namespace std;const int MaxN = 1 << 30;
const double eps = 1e-8;
const double PI = 3.141592654;int r;
int x[5], y[5];int sgn(double d)
{if(d > eps) return 1;if(d < -eps) return -1;return 0;
}double mymax(double a, double b)
{if(sgn(a - b) > 0) return a;return b;
}double getAngle(int x1, int y1, int x2, int y2)
{double up = (double)(x1 * x2 + y1 * y2);double down = sqrt(double(x1 * x1 + y1 * y1)) * sqrt(double(x2 * x2 + y2 * y2));return acos(up / down) * 180.0 / PI;
}double gao()
{double ang1 = getAngle(x[0], y[0], x[1], y[1]);double ang2 = getAngle(x[0], y[0], x[2], y[2]);double ang3 = getAngle(x[1], y[1], x[2], y[2]);return mymax(mymax(ang1, ang2), ang3);
}int main()
{int t;wh(t) {while(t--) {scanf("%d", &r);rep(i, 3) scanf("%d%d", &x[i], &y[i]);bool flag = false;rep(i, 3) if(0 == x[i] && 0 == y[i]) {puts("No");flag = true;break;}if(flag) continue;double max_angle = gao();if(sgn(max_angle - 120.0) < 0) puts("No");else puts("Yes"); }}return 0;
}
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