#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define ld long double
#define ull unsigned long long
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define rep(i, b) for(int i = 0; i < (b); ++i)
#define rep2(i,a,b) for(int i = a; i <= (b); ++i)
#define rep3(i,a,b,c) for(int i = a; i <= (b); i+=c)
#define count_bits(x) __builtin_popcountll((x))
#define all(x) (x).begin(),(x).end()
#define siz(x) (int)(x).size()
#define forall(it,x) for(auto& it:(x))
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//mt19937 mt;void random_start(){mt.seed(chrono::time_point_cast<chrono::milliseconds>(chrono::high_resolution_clock::now()).time_since_epoch().count());}
//ll los(ll a, ll b) {return a + (mt() % (b-a+1));}
const int INF = 1e9+50;
const ll INF_L = 1e18+40;
const ll MOD = 1e9+7;
ll pref[20001];
bool can_solve[50][20001];
int nxt_zero[50][20001];
ll best[51][51];
ll dp[20001][51];
vector<pii> events[20003];
int main()
{
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//random_start();
int n,t,s;
cin >> n >> t >> s;
rep2(i,1,t)
{
ll x;
cin >> x;
pref[i] = pref[i-1]+x;
}
rep(i,n)
{
can_solve[i][0] = 1;
rep2(j,1,t)
{
char z;
cin >> z;
can_solve[i][j] = z-'0';
}
int pop = t+1;
for(int j = t; j >= 0; j--)
{
if(!can_solve[i][j]) pop = j;
nxt_zero[i][j] = pop;
}
}
rep2(i,0,t-1)
{
vi x;
rep(j,n) x.pb(nxt_zero[j][i+1]);
sort(all(x));
rep(j,n) events[x[j]].pb({i,n-j-1});
}
rep2(i,0,t) rep2(j,0,s) dp[i][j] = 1e18;
rep2(i,0,s) rep2(j,0,n) best[i][j] = 1e18;
best[0][n] = 0;
dp[0][0] = 0;
rep2(i,1,t)
{
forall(it,events[i])
{
rep(j,s+1) best[j][it.ss] = min(best[j][it.ss],dp[it.ff][j]-pref[it.ff]*it.ss);
}
rep2(j,1,s)
{
rep(k,n+1)
{
dp[i][j] = min(dp[i][j],best[j-1][k]+pref[i]*k);
}
}
rep2(j,1,s) best[j][n] = min(best[j][n],dp[i][j]-pref[i]*n);
}
rep2(i,1,s) cout << dp[t][i] << "\n";
}
