1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
//solve O(nm(k^2))
//
//licze na 3 punkty
//
//////////////////////////////////////////////////
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define tak cout<<"TAK\n"
#define yes cout<<"YES\n"
#define nie cout<<"NIE\n"
#define no cout<<"NO\n"
using ll = long long;
using ld = long double;
using pll = pair<ll,ll>;
using pdd = pair<ld,ld>;
#define vec vector
#define all(x) (x).begin(), (x).end()
#define sz(x) (x).size()
#define linijki ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mapa map<ll,ll>
#define umapa unordered_map<ll,ll>
#define se set<ll>
#define ff first
#define ss second
#define str string
const ll alph = 26;
const ll BIN6 = (1LL << 20);
const ll BIN9 = (1LL << 30);
const ll BIN18 = (1LL << 60);
const ll INF5 = 100000;
const ll INF6 = 1000000;
const ll INF9 = 1000000000;
const ll INF18 = 1000000000000000000;
const ld DINF5 = 1e5L;
const ld DINF6 = 1e6L;
const ld DINF9 = 1e9L;
const ll MOD = 1000000007;

int main() {
    linijki;
    
    ll n,m,k;
    cin >> n>>m>>k;
    vec<ll> dp(k+10,(INF18*(-1)));
    dp[0]=0;
    for(ll i=0;i<n;++i){
        vec<ll> v(m);
        for(ll j=0;j<m;++j){
            cin>>v[j];
        }
        vec<ll> pref(m+1,0);
        for(ll j=1;j<=m;++j){
            pref[j]=pref[j-1]+v[j-1];
        }        
        vec<ll> pom=dp;
        for(ll i=0;i<=k;++i){
            for(ll j=1;(j<=m and (i+j) <=k);++j){
                pom[(i+j)]=max(pom[(i+j)],(dp[i]+pref[j]));
            }
        }
        dp=pom;
    }
    cout<<dp[k];
    
    return 0;
}