题解:

在圆上点三个点时,除圆上三个交点外,圆内没有交点;在圆上点四个点时,除圆上四个交点外,圆内出现了一个交点,因此,在N个点中每四个点便可以在圆内产生一个交点,因此N个点在圆内形成的点的个数为CN4,总的交点数就是CN4+N

 #pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
using namespace std;
#define ull unsigned long long int
#define eps 1e-10
#define MOD 1000000007
#define N 1000000
#define inf 1e12
ull n;
int main()
{
int t;
scanf("%d",&t);
while(t--){
scanf("%I64u",&n);
ull ans=;
if(n>=)
ans=ans+n*(n-)/*(n-)/*(n-)/;
ans+=n;
printf("%I64u\n",ans);
}
return ;
}

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