P6521 [CEOI2010 day2] pin

Solution

\(\texttt{E}\color{red}{\texttt{ricQian}}\) 很有精神！\(\color{white}{114514}\)

我们都要像他一样有精神！\(\color{white}{114514}\)

\(\texttt{E}\color{red}{\texttt{ricQian}}\) 说你可以设一个 \(d\) 表示恰好某些位置相同的对数，\(p\) 表示至少某些位置相同的对数，\(D\) 表示恰好几个位置相同的对数，\(P\) 表示至少几个位置相同的对数。\(\color{white}{114514}\)

\(\texttt{E}\color{red}{\texttt{ricQian}}\) 说 \(P\) 和 \(p\) 都可以直接求。于是我们来考虑 \(D\) 和 \(d\) 的求法。\(\color{white}{114514}\)

\(\texttt{E}\color{red}{\texttt{ricQian}}\) 觉得你很逊，于是他来帮你颓式子：\(\color{white}{114514}\)

\[\begin{aligned} D_4&=P_4\\ D_3&=d_{123}+d_{124}+d_{134}+d_{234}\\ &=(p_{123}-p_{1234})+(p_{124}-p_{1234})+(p_{134}-p_{1234})+(p_{234}-p_{1234})\\ &=P_3-4D_4\\ D_2&=d_{12}+d_{13}+d_{14}+d_{23}+d_{24}+d_{34}\\ &=(p_{12}-d_{123}-d_{124}-d_{1234})+(p_{13}-d_{123}-d_{134}-d_{1234})+(p_{14}-d_{124}-d_{134}-d_{1234})+(p_{23}-d_{123}-d_{234}-d_{1234})+(p_{24}-d_{124}-d_{234}-d_{1234})+(p_{34}-d_{134}-d_{234}-d_{1234})\\ &=P_2-3D_3-6D_4\\ D_1&=d_1+d_2+d_3+d_4\\ &=(p_1-d_{12}-d_{13}-d_{14}-d_{123}-d_{124}-d_{134}-d_{1234})+(p_2-d_{12}-d_{23}-d_{24}-d_{123}-d_{124}-d_{234}-d_{1234})+(p_3-d_{13}-d_{23}-d_{34}-d_{123}-d_{234}-d_{134}-d_{1234})+(p_4-d_{14}-d_{24}-d_{34}-d_{124}-d_{134}-d_{234}-d_{1234})\\ &=P_1-2D_2-3D_3-4D_4 \end{aligned} \]

哇！这真是太有精神了吧！真的没有再比这样的事情更有精神了！\(\color{white}{114514}\)

我们的 \(\texttt{E}\color{red}{\texttt{ricQian}}\) 真是太厉害了！大家快来膜拜 \(\texttt{E}\color{red}{\texttt{ricQian}}\)！！！\(\color{white}{114514}\)

Code

#define int long long
using namespace std;
const int MAXN=5e4+10;
map<string,int> cnt;
string s[MAXN];
int P[MAXN],D[MAXN];
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	int n,d;cin>>n>>d;
	rep(i,1,n) cin>>s[i];
	rep(i,1,15){
		int ppc=__builtin_popcount(i);
		cnt.clear();
		rep(j,1,n){
			string t;
			rep(k,0,3)
				if(i&(1<<k))
					t+=s[j][k];
			P[ppc]+=(cnt[t]++);
		}
	}D[4]=P[4];
	D[3]=P[3]-4*D[4];
	D[2]=P[2]-3*D[3]-6*D[4];
	D[1]=P[1]-2*D[2]-3*D[3]-4*D[4];
	if(d==4) cout<<n*(n-1)/2-D[1]-D[2]-D[3]-D[4]<<'\n';
	else cout<<D[4-d]<<'\n';
	return 0;
}