题意:给定二维平面上的n个整点,问它们是否都在正n边形的定点上

n<=100,abs(x[i]),abs(y[i])<=1e4

思路:队友做的,抱大腿

可以发现只有n=4时顶点有可能都是整点,判一下对角线与边长就行

我与队友互演,WA了5发……

 #include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn=1e3+;
struct node{
int x,y;
}a[maxn];
int dis(int i,int j)
{
return (a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
}
if(n!=)
{
printf("NO\n");
}
else
{
int g=;
int d=;
for(int i=;i<=;i++)
{
if(dis(,i)>d)
{
g=i;
d=dis(,i);
}
}
if(g!=)
swap(a[],a[g]);
int f=;
if(dis(,)==dis(,)&&dis(,)==dis(,)&&dis(,)==dis(,)&&dis(,)==dis(,))
f=;
if(f)
printf("YES\n");
else
printf("NO\n");
}
} }

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