解析
- 法一:每 4 个顶点会有一个交点,答案为 $ C_n^4 $
- 法二:递推做差找规律
Code
法一
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
#define ULL unsigned long long
using namespace std;
ULL n;
int main()
{
scanf("%llu",&n);
printf("%llu
",n*(n-1)/2*(n-2)/3*(n-3)/4);
return 0;
}
法二
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
const int N=1e5+5;
int n;
LL a[N],b[N],c[N];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
a[i]=a[i-1]+i;
b[i]=b[i-1]+a[i];
c[i]=c[i-1]+b[i];
}
printf("%lld
",c[n-3]);
return 0;
}