http://acm.hdu.edu.cn/showproblem.php?pid=2036
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6885 Accepted Submission(s): 3417
多边形公式:0.5*|x1*y2-y1*x2+x2*y3-y2*x3+……+xn*y1-yn*x1|
绝对值函数:int型:abs(int x),float型 fabs(float x)
#include<stdio.h> void main() { int n,x[3],y[3]; double s; while(scanf("%d",&n)!=EOF) { if(n==0) break; if(n>=3&&n<=100) { s=0; scanf("%d%d",&x[0],&y[0]); x[2]=x[0]; y[2]=y[0]; while(--n) { scanf("%d%d",&x[1],&y[1]); s+=x[0]*y[1]-x[1]*y[0]; x[0]=x[1]; y[0]=y[1]; } s+=x[0]*y[2]-x[2]*y[0]; printf("%.1lf\n",s/2); } } } //多边形求面积公式:S = 0.5 * ( (x0*y1-x1*y0) + (x1*y2-x2*y1) + ... + (xn*y0-x0*yn) //假设只有两个点的话,那么最后相乘为a[2][0]*a[0][1]-a[0][0]*a[2][1] //这一项多出来了,因为根本不存在a[2][0],a[2][1],但是只要a[2][0]=a[0][0],a[2][1]=a[0][1],那么上式的差为0,不影响结果! #include<iostream> #include<iomanip> #include<math.h> using namespace std; int main() { int n,a[100][2]; while(cin>>n&&n!=0) { if(n>=3&&n<=100) { int i; double sum=0; for(i=0;i<n;++i) cin>>a[i][0]>>a[i][1]; a[n][0]=a[0][0]; a[n][1]=a[0][1]; for(i=1;i<=n;++i) sum+=a[i-1][0]*a[i][1]-a[i][0]*a[i-1][1]; cout<<setiosflags(ios::fixed)<<setprecision(1)<<sum/2<<endl; } } return 0; }