#include "stdafx.h"
#include
#include
#include
#include
using namespace std;
struct point {int x, y;
}P[100];
int n;//顶点个数int judge() {double angle = 0;//较小角的和point a, b;a.x = P[1].x - P[0].x; a.y = P[1].y - P[0].y;b.x = P[n - 1].x - P[0].x; b.y = P[n - 1].y - P[0].y;angle += acos((a.x*b.x + a.y*b.y) / (sqrt(pow(a.x, 2) + pow(a.y, 2))*sqrt(pow(b.x, 2) + pow(b.y, 2))));//第一个角角度求解a.x = P[0].x - P[n - 1].x; a.y = P[0].y - P[n - 1].y;b.x = P[n - 2].x - P[n - 1].x; b.y = P[n - 2].y - P[n - 1].y;angle += acos((a.x*b.x + a.y*b.y) / (sqrt(pow(a.x, 2) + pow(a.y, 2))*sqrt(pow(b.x, 2) + pow(b.y, 2))));//最后一个角角度求解 for (int i = 1; i < n - 1; i++) {a.x = P[i + 1].x - P[i].x; a.y = P[i + 1].y - P[i].y;b.x = P[i - 1].x - P[i].x; b.y = P[i - 1].y - P[i].y;angle += acos((a.x*b.x + a.y*b.y) / (sqrt(pow(a.x, 2) + pow(a.y, 2))*sqrt(pow(b.x, 2) + pow(b.y, 2))));}//计算所有较小角的和 若等于（n-2）pi 则凸cout << angle << endl;cout << (n - 2)*acos(-1) << endl;if (angle == (n - 2)*acos(-1))printf("凸");else printf("凹");return 0;
}int main() {scanf_s("%d", &n);for (int i = 0; i < n; i++)scanf_s("%d%d", &P[i].x, &P[i].y);judge();system("pause");
} 每个顶点的较小角（内角或外角）相加，若等于（顶点数-2）*180度，这为凸n边形，若小于，则为凹n边形。
可以用向量求cos值，再用反三角函数就角度，得出来的就是较小角（内角或者外角）。