#include <stdio.h>
struct Node
{
int x, y;
}node[105];
//(x0,y0), (x1,y1), (x2, y2)是逆时针返回值为正,顺时针返回值为负
double proportion(Node a, Node b, Node c)
{
return 0.5*(a.x*b.y + b.x*c.y + c.x*a.y - c.x*b.y - b.x*a.y - a.x*c.y);
}
int main()
{
int n;
while(scanf("%d", &n)) {
if(n == 0) break;
for(int i = 0; i < n; i++)
scanf("%d%d", &node[i].x, &node[i].y);
double sum = 0.0;
for(int i = 2; i < n; i++) {
sum += proportion(node[0], node[i - 1], node[i]);
}
printf("%.1lf\n", sum);
}
return 0;
}
总结
这道题考察多边形的面积,可怜我的数学,没有知识基础呀。
S
=
0.5
∗
∑
i
=
1
n
−
1
∣
1
x
0
y
0
1
x
i
y
i
1
x
i
+
1
y
i
+
1
∣
S = 0.5 \ast \sum_{i = 1}^{n-1}\left | \begin{array} {ccc} 1 & x_0 & y_0 \\ 1 & x_i & y_i \\ 1 & x_{i+1} & y_{i+1} \end {array} \right |
S=0.5∗i=1∑n−1∣∣∣∣∣∣111x0xixi+1y0yiyi+1∣∣∣∣∣∣
其中S在算完后需要再取一个绝对值
理论基础在这:http://blog.sina.com.cn/s/blog_7c50857d01014z0g.html