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.
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
int trans(char* c)
{
int len=strlen(c),v=0;
for(int i=0;i<len;i++) v=2*v+(c[i]-'0');
return v;
}
const int maxn=1e5+5;
char seq[20];
int cost[20],v[maxn];
const int ub=(1<<12)+5;
int cnt[ub][200];
int n,m,q;
/*
題目大意:題目意思很煩，反正就是給一堆二進制串集合，
然後給定q個查詢，集合個數和查詢數量都很大。
查詢和集合數量雖然多但真正的狀态很少，
是以可以通過預處理計數的方式，
這道題還是蠻考研基本功的，雖然沒有什麼特别困難的思維。
對位運算的掌握，對同或的掌握等等，差距展現在速度上。
簡單dp預處理後求個字首和即可。
*/
int main()
{
scanf("%d%d%d",&n,&m,&q);
for(int i=n-1;i>=0;i--) scanf("%d",&cost[i]);
for(int i=0;i<m;i++)
{
scanf("%s",seq);
v[trans(seq)]++;///計數數組
}
for(int i=0;i<(1<<n);i++)
{
for(int j=0;j<(1<<n);j++)
{
int sta=~(i^j),cst=0;
for(int k=0;k<n;k++) if(sta&(1<<k)) cst+=cost[k];
if(cst<=100) cnt[i][cst]+=v[j];
///for(int k=cst;k<=100;k++) cnt[i][k]+=v[j];///逾時的做法
}
for(int j=1;j<=100;j++) cnt[i][j]+=cnt[i][j-1];
}
char c[maxn];int tp;
for(int i=0;i<q;i++)
{
scanf("%s%d",c,&tp);
printf("%d\n",cnt[trans(c)][tp]);
}
return 0;
}