【比赛】HNOI2018 寻宝游戏

考试的时候就拿了30points滚粗了

听说myy对这题的倒推做法很无奈，官方题解在此

正解思路真的很巧妙，也说的很清楚了

就是分别考虑每一位，会发现题解中的那个性质，然后把询问的二进制数按照排序后的位置放好，如果有1在0后面，那就肯定puts("0")，否则根据那个式子，每一位都要满足那个条件，最后范围不断缩小，就变成了两个数的差

#include<bits/stdc++.h>

#define ui unsigned int

#define ll long long

#define db double

#define ld long double

#define ull unsigned long long

const int MAXN=1000+10,MAXM=5000+10,Mod=1e9+7;

ll total;

int n,m,q,r[MAXM];

char bin[MAXM];

template<typename T> inline void read(T &x)

{

	T data=0,w=1;

	char ch=0;

	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();

	if(ch=='-')w=-1,ch=getchar();

	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();

	x=data*w;

}

template<typename T> inline void write(T x,char ch='\0')

{

	if(x<0)putchar('-'),x=-x;

	if(x>9)write(x/10);

	putchar(x%10+'0');

	if(ch!='\0')putchar(ch);

}

template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}

template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}

template<typename T> inline T min(T x,T y){return x<y?x:y;}

template<typename T> inline T max(T x,T y){return x>y?x:y;}

inline ll qexp(ll a,ll b)

{

	ll res=1;

	while(b)

	{

		if(b&1)res=res*a%Mod;

		a=a*a%Mod;

		b>>=1;

	}

	return res;

}

struct data{

	int b[MAXN],id;

	ll ans;

	data(){ans=-1;}

	inline bool operator < (const data &A) const {

		for(register int i=1;i<=n;++i)

			if(b[i]!=A.b[i])return b[i]>A.b[i];

		return false;

	};

	inline ll value()

	{

		if(~ans)return ans;

		ans=0;

		for(register int i=1;i<=n;++i)(ans+=b[i]*qexp(2,n-i))%=Mod;

		return ans;

	}

};

data infin[MAXM];

int main()

{

	freopen("hunt.in","r",stdin);

	freopen("hunt.out","w",stdout);

	read(n);read(m);read(q);

	for(register int i=1;i<=n;++i)

	{

		scanf("%s",bin+1);

		for(register int j=1;j<=m;++j)infin[j].b[n-i+1]=bin[j]-'0';

	}

	for(register int i=1;i<=m;++i)infin[i].id=i;

	total=qexp(2,n);

	std::sort(infin+1,infin+m+1);

	while(q--)

	{

		scanf("%s",bin+1);

		for(register int i=1;i<=m;++i)r[i]=bin[infin[i].id]-'0';

		int p0=m+1,p1=0;

		for(register int i=m;i>=1;--i)

			if(!r[i])p0=i;

		for(register int i=1;i<=m;++i)

			if(r[i])p1=i;

		if(p0<p1)puts("0");

		else if(p0==1)write((total-infin[1].value()+Mod)%Mod,'\n');

		else if(p0==m+1)write(infin[m].value(),'\n');

		else write((infin[p0-1].value()-infin[p0].value()+Mod)%Mod,'\n');

	}

	return 0;

}