[SDOI2013]spring

Description

Input

Output

Sample Input
3 3
1 2 3 4 5 6
1 2 3 0 0 0
0 0 0 4 5 6

Sample Output
2

HINT

---

容斥部分其实很好想，我们考虑枚举有哪些泉区的泉水可能相同，答案即为(Ans=sumlimits_{i=k}^6(-1)^{i-k} imes inom{i}{k} imes work(i))

考虑如何求work？我们可以(2^6)枚举考虑哪些泉区，然后对其进行hash，记录下相同hash值的cnt，每次枚举的答案即为(sumlimits_{i=1}^kdfrac{cnt_i(cnt_i+1)}{2})

不要用map，不要用map，不要用map，请自己手打hash，不要问我怎么知道的

/*problem from Wolfycz*/
#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-');
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5,Mod=5249857;
int C[10][10],v[10][N+10],cnt[10];
int n,k,limit;
struct S1{//hash链
	int pre[N+10],now[Mod+10],val[N+10],tot;
	ull child[N+10];
	void clear(){
		tot=0;
		memset(now,0,sizeof(now));
		memset(val,0,sizeof(val));
	}
	void insert(ull x){
		int Now=x%Mod,p; ull son;
		for (p=now[Now],son=child[p];p;p=pre[p],son=child[p]){
			if (son==x){
				val[p]++;
				return;
			}
		}
		pre[++tot]=now[Now],now[Now]=tot,child[tot]=x,val[tot]=1;
	}
}Hash;
void Extract(int sta,int &len){for (int i=1;i<=6;++i)	cnt[i]=sta&1,len+=sta&1,sta>>=1;}
ll work(int len){
	Hash.clear();
	for (int i=1;i<=n;++i){
		ull tmp=0;
		for (int j=1;j<=6;j++)	if (cnt[j])	tmp=tmp*limit+v[j][i];
		Hash.insert(tmp);
	}
	ll res=0;
	for (int i=1;i<=Hash.tot;i++)	res+=1ll*Hash.val[i]*(Hash.val[i]-1)/2;
	return res;
}
bool check(int x){
	for (int i=2;i*i<=x;i++)	if (x%i==0)	return 0;
	return 1;
}
void prepare(){//得到随机膜数
	srand(time(0));
	int p=1e9;
	while (true){
		int x=1ll*rand()*rand()%p;
		if (x<1000000)	continue;
		if (check(x)){
			limit=x;
			break;
		}
	}
}
int main(){
	prepare();
	n=read(),k=read(); ll Ans=0;
	for (int i=1;i<=n;++i)	for (int j=1;j<=6;++j)	v[j][i]=read();
	for (int i=0;i<=6;++i){
		C[i][0]=1;
		for (int j=1;j<=i;++j)	C[i][j]=C[i-1][j-1]+C[i-1][j];
	}
	for (int sta=0;sta<1<<6;++sta){
		int len=0;
		Extract(sta,len);
		if (len<k)	continue;
		ll res=work(len)*C[len][k];
		(len-k)&1?Ans-=res:Ans+=res;
	} 	
	printf("%lld
",Ans);
	return 0;
}