Area POJ - 1654
Area
题意：就是让你求多边形面积
思路：利用多边形面积公式即可，输出格式第一次见。。。wa了好几发。。
1 // 2 // Created by HJYL on 2020/2/3. 3 // 4 #include<iostream> 5 #include<cstring> 6 #include<cstdio> 7 #include<cstring> 8 #include<cmath> 9 #include<algorithm> 10 using namespace std; 11 const double eps=1e-8; 12 const int maxn=1e6+5; 13 struct Point{ 14 int x,y; 15 Point(int x=0,int y=0):x(x),y(y){} 16 Point operator - (Point const &b)const 17 { 18 return Point(x-b.x ,y-b.y); 19 } 20 bool operator < (Point const &c)const{ 21 if(x==c.x) 22 return y<c.y; 23 return x<c.x; 24 } 25 }p[maxn]; 26 double Cross(Point A,Point B) 27 { 28 return (A.x*B.y)-(A.y*B.x); 29 } 30 long long PolygonArea(Point* p, int n) {//p为端点集合，n为端点个数 31 long long s = 0; 32 for (int i = 1; i < n - 1; ++i) 33 s += Cross(p[i] - p[0], p[i + 1] - p[0]); 34 return s<0?-s:s; 35 } 36 char s[maxn]; 37 const int dx[]={0,-1,0,1,-1,0,1,-1,0,1}; 38 const int dy[]={0,-1,-1,-1,0,0,0,1,1,1}; 39 int n; 40 int main() 41 { 42 while(~scanf("%d",&n)) 43 { 44 while(n--) 45 { 46 getchar(); 47 scanf("%s",s); 48 int l=strlen(s); 49 p[0].x=0,p[0].y=0; 50 for(int i=1;i<l;i++) { 51 p[i].x = p[i - 1].x +dx[s[i]-'0'], p[i].y = p[i - 1].y +dy[s[i]-'0']; 52 } 53 long long ss=PolygonArea(p,l); 54 if(ss%2==0) printf("%lld\n",ss/2); 55 else printf("%lld.5\n",ss/2); 56 } 57 } 58 return 0; 59 }
