2n条直线划分平面总数为2n^2+n+1;
除去4n条边和2n个顶点后变成n条折线,由欧拉公式可得面数减少2n,即2n^2-n+1;
#include<math.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
int t,n;
long long f[20001];
f[0]=1;
for(int i=1; i<20001; i++)
f[i]=f[i-1]+i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("%I64d\n",f[2*n]-2*n);
}
return 0;
}
原文地址:http://blog.csdn.net/loolu5/article/details/41264257