codeforces 1017 problem D. The Wu

D. The Wu

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Childan is making up a legendary story and trying to sell his forgery — a necklace with a strong sense of "Wu" to the Kasouras. But Mr. Kasoura is challenging the truth of Childan's story. So he is going to ask a few questions about Childan's so-called "personal treasure" necklace.

This "personal treasure" is a multiset SS of mm "01-strings".

A "01-string" is a string that contains nn characters "0" and "1". For example, if n=4n=4, strings "0110", "0000", and "1110" are "01-strings", but "00110" (there are 55 characters, not 44) and "zero" (unallowed characters) are not.

Note that the multiset SS can contain equal elements.

Frequently, Mr. Kasoura will provide a "01-string" tt and ask Childan how many strings ss are in the multiset SS such that the "Wu" value of the pair (s,t)(s,t) is not greater than kk.

Mrs. Kasoura and Mr. Kasoura think that if si=tisi=ti (1≤i≤n1≤i≤n) then the "Wu" value of the character pair equals to wiwi, otherwise 00. The "Wu" value of the "01-string" pair is the sum of the "Wu" values of every character pair. Note that the length of every "01-string" is equal to nn.

For example, if w=[4,5,3,6]w=[4,5,3,6], "Wu" of ("1001", "1100") is 77 because these strings have equal characters only on the first and third positions, so w1+w3=4+3=7w1+w3=4+3=7.

You need to help Childan to answer Mr. Kasoura's queries. That is to find the number of strings in the multiset SS such that the "Wu" value of the pair is not greater than kk.

Input

The first line contains three integers nn, mm, and qq (1≤n≤121≤n≤12, 1≤q,m≤5⋅1051≤q,m≤5⋅105) — the length of the "01-strings", the size of the multiset SS, and the number of queries.

The second line contains nn integers w1,w2,…,wnw1,w2,…,wn (0≤wi≤1000≤wi≤100) — the value of the ii-th caracter.

Each of the next mm lines contains the "01-string" ss of length nn — the string in the multiset SS.

Each of the next qq lines contains the "01-string" tt of length nn and integer kk (0≤k≤1000≤k≤100) — the query.

Output

For each query, print the answer for this query.

Examples

input

Copy

2 4 5
40 20
01
01
10
11
00 20
00 40
11 20
11 40
11 60

output

Copy

2
4
2
3
4

input

Copy

1 2 4
100
0
1
0 0
0 100
1 0
1 100

output

Copy

1
2
1
2

Note

In the first example, we can get:

"Wu" of ("01", "00") is 4040.

"Wu" of ("10", "00") is 2020.

"Wu" of ("11", "00") is 00.

"Wu" of ("01", "11") is 2020.

"Wu" of ("10", "11") is 4040.

"Wu" of ("11", "11") is 6060.

In the first query, pairs ("11", "00") and ("10", "00") satisfy the condition since their "Wu" is not greater than 2020.

In the second query, all strings satisfy the condition.

In the third query, pairs ("01", "11") and ("01", "11") satisfy the condition. Note that since there are two "01" strings in the multiset, the answer is 22, not 11.

In the fourth query, since kk was increased, pair ("10", "11") satisfies the condition too.

In the fifth query, since kk was increased, pair ("11", "11") satisfies the condition too.

 

题意：给出n,m,q，有m个01串，每个串的长度为n，给出n个值，值ai表示01串第i个位置的权值（从左至右）如果另一个串的对应位置的元素和这个串的元素相同，那么就可以获得这个权值。给出q次查询，每次查询给出一个长为n的01串，和一个值k（0<=k<=100） 。问这个串和给出的m个串进行操作，获得权值不大于k的个数是多少。

 

思路：m和q都很大，所以，我们只能考虑从n入手了，因为查询的次数很多，所以需要预处理。由于是01串，所以我们可以考虑进行状态压缩，将一个01串压缩成一个10进制的数字，n最大为12，那么压缩为10进制的数的话，最大也就4096.这样，我们就可以枚举所有的状态了。记录给出的m串每个串出现的次数（先压缩再记录）这样的话，最坏的情况也只有4096种了。然后两个for循环，枚举所有的状态，并记录对应状态下获得权值的个数。最后查询的时候累加一下就好了。

 

#include<iostream>
#include<cstdio>
using namespace std;
const int Max=5e5+10;
int check[15],a[Max],num[5000][105],mmp[5000];
int main()
{
    char x[15];
    int n,m,q,tmp,u,v,cnt=1,sum;
    scanf("%d%d%d",&n,&m,&q);
    for(int i=1;i<=n;++i) scanf("%d",&check[i]);
    for(int i=1;i<=m;++i){
        tmp=0;
        scanf("%s",x);
        for(int j=0;j<n;++j)
            tmp=(tmp<<1)+x[j]-'0';//进行状态压缩，二进制转换为十进制  
        if(!mmp[tmp]) a[cnt++]=tmp;
        ++mmp[tmp];//记录每种状态的个数
    }
    for(int i=0;i<(1<<n);++i){
        for(int j=1;j<cnt;++j){
            u=i,v=a[j];sum=0;
            for(int k=n;k>=1;--k){
                if((u&1)==(v&1)) sum+=check[k];//u&1相当于u%2
                if(sum>100) break;//由于k小于100，所以大于100不记录
                u>>=1,v>>=1;
            }
            if(sum<=100) num[i][sum]+=mmp[a[j]];
        }
    }
    int xx,ans;
    for(int i=0;i<q;i++){
        ans=tmp=0;
        scanf("%s",x);
        for(int j=0;j<n;j++)
            tmp=(tmp<<1)+x[j]-'0';
        scanf("%d",&xx);
        for(int j=0;j<=xx;j++)
            ans+=num[tmp][j];//累加所有的权值的个数
        printf("%d\n",ans);
    }
    return 0;
}