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题意:从原点出发向八个方向走,所给数字串每个数字代表一个方向,终点与原点连线,求所得多边形面积;
思路:(性质)共起点的两向量叉积的一半为两向量围成三角形的面积。以此计算每条边首尾两个向量的叉积,求和,除二;
#include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const double epsi=1e-10; const double pi=acos(-1.0); const int maxn=100005; inline int sign(const double &x){ if(x>epsi) return 1; if(x<-epsi) return -1; return 0; } struct point{ long long x,y; point(double xx,double yy):x(xx),y(yy) {} point operator +(const point &op2) const{ return point(x+op2.x,y+op2.y); } long long operator ^(const point &op2) const{ return x*op2.y-y*op2.x; } }; int main() { int t,i,j,k; string s; long long ans; scanf("%d",&t); while(t--){ cin>>s; ans=0; point p=point(0,0),p1=point(0,0); for(int i=0;i<s.size();i++){ if(s[i]==‘1‘) p1=p+point(-1,-1); if(s[i]==‘2‘) p1=p+point(0,-1); if(s[i]==‘3‘) p1=p+point(1,-1); if(s[i]==‘4‘) p1=p+point(-1,0); if(s[i]==‘5‘) p1=p+point(0,0); if(s[i]==‘6‘) p1=p+point(1,0); if(s[i]==‘7‘) p1=p+point(-1,1); if(s[i]==‘8‘) p1=p+point(0,1); if(s[i]==‘9‘) p1=p+point(1,1); ans+=p^p1; p=p1; } if(ans<0) ans=-ans; printf("%lld",ans/2); if(ans%2) printf(".5"); printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/dashuzhilin/p/4539554.html