making lines memory nts type next .org efi math.h

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 $\mathrm{SS\; of\; mm\; "01-strings".}$

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

Note that the multiset

SS can contain equal elements.

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

Mrs. Kasoura and Mr. Kasoura think that if ${\mathrm{si=tisi=ti\; (1\le i\le n1\le i\le n)\; then\; the\; "Wu"\; value\; of\; the\; character\; pair\; equals\; to}}^{}$

wiwi, otherwise $\mathrm{00.\; The\; "Wu"\; value\; of\; the\; "01-string"\; pair\; is\; the\; sum\; of\; the\; "Wu"\; values\; of\; every\; character\; pair.\; Note\; that\; the\; length\; of\; every\; "01-string"\; is\; equal\; to\; nn.}$

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

You need to help Childan to answer Mr. Kasoura‘s queries. That is to find the number of strings in the multiset $\mathrm{SS\; such\; that\; the\; "Wu"\; value\; of\; the\; pair\; is\; not\; greater\; than\; kk.}$

Input

The first line contains three integers $\mathrm{nn,\; mm,\; and\; qq\; (1\le n\le 121\le n\le 12,\; 1\le q,m\le 5\cdot 1051\le q,m\le 5\cdot 105)\; —\; the\; length\; of\; the\; "01-strings",\; the\; size\; of\; the\; multiset\; SS,\; and\; the\; number\; of\; queries.}$

The second line contains $\mathrm{nn\; integers\; w1,w2,\dots ,wnw1,w2,\dots ,wn\; (0\le wi\le 1000\le wi\le 100)\; —\; the\; value\; of\; the\; ii-th\; caracter.}$

Each of the next $\mathrm{mm\; lines\; contains\; the\; "01-string"\; ss\; of\; length\; nn\; —\; the\; string\; in\; the\; multiset\; SS.}$

Each of the next $\mathrm{qq\; lines\; contains\; the\; "01-string"\; tt\; of\; length\; nn\; and\; integer\; kk\; (0\le k\le 1000\le k\le 100)\; —\; the\; query.}$

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 $\mathrm{22,\; not\; 11.}$

In the fourth query, since $\mathrm{kk\; was\; increased,\; pair\; ("10",\; "11")\; satisfies\; the\; condition\; too.}$

In the fifth query, since $\mathrm{kk\; was\; increased,\; pair\; ("11",\; "11")\; satisfies\; the\; condition\; too.}$

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define MAX 500005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
typedef long long ll;

int a[MAX];
char s[MAX][20];
char ss[20];
ll dp[(1<<12)+5][105];
int b[(1<<12)+5];

int main()
{
    int n,m,q,x,i,j,k;
    scanf("%d%d%d",&n,&m,&q);
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    for(i=0;i<m;i++){
        scanf(" %s",s[i]);
    }
    for(i=0;i<m;i++){
        ll S=0;
        for(j=0;j<n;j++){
            if(s[i][j]==‘1‘) S|=1<<j;
        }
        b[S]++;
    }
    for(i=0;i<(1<<n);i++){
        for(j=0;j<(1<<n);j++){
            if(!b[j]) continue;
            int sum=0;
            for(k=0;k<n;k++){
                if((i&(1<<k))==(j&(1<<k))) sum+=a[k];
            }
            if(sum>100) continue;
            dp[i][sum]+=b[j];
        }
    }
    
    while(q--){
        scanf(" %s %d",ss,&x);
        ll S=0;
        for(i=0;i<n;i++){
            if(ss[i]==‘1‘) S|=1<<i;
        }
        ll ans=0;
        for(i=0;i<=x;i++){
            ans+=dp[S][i];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

Codeforces Round #502 D. The Wu（狀壓預處理）