【codeforces 514C】Watto and Mechanism(字符串hash)

【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("
%.2lf sec 
", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}