【题目链接】:http://codeforces.com/contest/514/problem/C

【题意】



给你n个字符串;

然后给你m个询问;->m个字符串

对于每一个询问字符串

你需要在n个字符串里面找到和它的长度相同,且只有一个位置的字符不同的字符串;

或者告知这是不存在的;

【题解】



字符串hash

因为只有3个字符

所以权值就为3^x;

找个大质数取模就好；

不够大就再大一点。。

然后对于每一位,尝试换成其他两个字母,看看存不存在;



【Number Of WA】



3



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define ps push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%lld",&x)

#define ref(x) scanf("%lf",&x)

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 6e5+100;

const LL MOD = 1e12+7;

int n,m;

LL p[N];

string s;

set<LL> myset;

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    p[0] = 1;

    rep1(i,1,N-2)

        p[i] = (p[i-1]*3)%MOD;

    rei(n),rei(m);

    rep1(i,1,n)

    {

        cin >> s;

        LL temp = 0;

        int len = s.size();

        rep1(j,0,len-1)

            temp = (temp*3+s[j]-'0')%MOD;

        myset.insert(temp);

    }

    rep1(i,1,m)

    {

        cin >> s;

        bool ok = false;

        int len = s.size();

        LL temp = 0,ttemp;

        rep1(j,0,len-1)

            temp = (temp*3+s[j]-'0')%MOD;

        rep1(j,0,len-1)

        {

            char t = s[j];

            for (char d='a';d<='c';d++)

            if (t!=d)

            {

                ttemp = (temp-(t-'0')*p[len-j-1])%MOD;

                if (ttemp<0) ttemp=(ttemp+MOD)%MOD;

                ttemp=(ttemp+(d-'0')*p[len-j-1])%MOD;

                if (myset.count(ttemp))

                {

                    ok = true;

                    break;

                }

            }

            if (ok) break;

        }

        if (ok)

            puts("YES");

        else

            puts("NO");

    }

    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);

    return 0;

}