【POJ1939】：Diplomatic License

题意

给出实例，求各个相邻两点的中点，注意点是x,y ,然后最后一个点是和第一个点做平均分配的

思路

如题意所说，稍微看下代码就明白咋回事了
注意的是必须要是保留六位的格式~

代码

#include <iostream>#include <iomanip>#define F fixed#define S setprecision(6)#define max 1000using namespace std;double x[max],y[max];int main(){    int n,i;    double pos1,pos2;    while(cin>>n){        cout<<n;        for(i=0;i<n;i++){            cin>>x[i]>>y[i];        }        for(i=0;i<n-1;i++){            pos1 =(x[i]+x[i+1])/2.0;            pos2 =(y[i]+y[i+1])/2.0;            cout<<" "<<F<<S<<pos1<<F<<S<<" "<<pos2;        }        pos1 = (x[n-1]+x[0])/2.0;        pos2 = (y[n-1]+y[0])/2.0;        cout<<" "<<F<<S<<pos1<<" "<<F<<S<<pos2<<endl;    }    return 0;}