then Vova has to repost all the pictures in the news feed. If k=2

then Vova can skip some pictures, but between every pair of consecutive pictures Vova has to repost at least one of them.

Your task is to calculate the maximum possible sum of values of reposted pictures if Vova follows conditions described above, or say that there is no way to satisfy all conditions.

The first line of the input contains three integers n,k

and x (1≤k,x≤n≤5000

) — the number of pictures in the news feed, the minimum length of segment with at least one repost in it and the number of pictures Vova is ready to repost.

The second line of the input contains n

integers a1,a2,…,an (1≤ai≤109), where ai is the beauty of the i

-th picture.

Print -1 if there is no way to repost some pictures to satisfy all the conditions in the problem statement.

Otherwise print one integer — the maximum sum of values of reposted pictures if Vova follows conditions described in the problem statement.

5 2 3
5 1 3 10 1

18

6 1 5
10 30 30 70 10 10

-1

4 3 1
1 100 1 1

100

題意 ： 給你 n 個數字，要求從中選出 x 個數字，但任意連續的長度為 k 的區間中必須至少選擇一個元素，詢問所選擇元素的最大的和是多少？

思路分析 ：

　　定義 dp[i][j] 表示前 i 個樹中選擇 j 個數的最大得分

程式碼示例 ：

n = 200

#define ll long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;

ll n, k, x;
ll a[205];
ll dp[205][205];

void solve() {
    memset(dp, -1*inf, sizeof(dp));
    //printf("%lld ++++\n", dp[0][0]);
    dp[0][0] = 0;
    for(ll i = 1; i <= n; i++){
        for(ll j = max(0ll, i-k); j <= i-1; j++){
            for(ll f = 1; f <= x; f++){
                dp[i][f] = max(dp[i][f], dp[j][f-1]+a[i]);
                //prllf("++++ %d %d %d %d\n", i, j, f, dp[i][f]);
            }
        }
    }    
    
    ll ans = -1;
    for(ll i = n; i > n-k; i--) ans = max(ans, dp[i][x]);
    printf("%lld\n", ans);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> k >> x;
    for(ll i = 1; i <= n; i++){
        scanf("%lld", &a[i]);        
    }
    solve();
    return 0;
}

 n = 5000

#define ll long long
const ll maxn = 1e6+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
typedef pair<ll, ll> P; // pos val
#define fi first
#define se second

ll n, k, x;
ll a[5005];
deque<P>que[5005];
ll dp[5005][5005];

void solve() {
    memset(dp, -1*inf, sizeof(dp));
    dp[0][0] = 0;
    que[0].push_back(P(0, 0));
    ll ans = -1;
    
    for(ll i = 1; i <= n; i++){
        ll pos = max(i-k, 0ll);
        for(ll j = 1; j <= x; j++){
            while(!que[j-1].empty() && que[j-1].front().fi < pos){
                que[j-1].pop_front();
            }
        }
        for(ll j = 1; j <= x; j++){
            if (!que[j-1].empty()) {
                ll p = que[j-1].front().fi;
                ll val = que[j-1].front().se;
                dp[i][j] = max(dp[i][j], dp[p][j-1]+a[i]);
                //printf("^^^^^^^^^^^ %lld %lld  %lld ++++  %lld %lld %lld\n", i, j, dp[i][j], p, j-1, dp[p][j-1]);
            }
        }
        for(ll j = 1; j <= x; j++){
            while(!que[j].empty() && dp[i][j] >= que[j].back().se) que[j].pop_back();
            if (dp[i][j] > 0) que[j].push_back(P(i, dp[i][j]));
            if (i > n-k) ans = max(ans, dp[i][j]);
        }
        //for(ll j = 1; j <= x; j++) {
            //printf("++++ %lld %lld %lld\n", i, j, dp[i][j]);
        //} 
    }    
    printf("%lld\n", ans);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n >> k >> x;
    for(ll i = 1; i <= n; i++){
        scanf("%lld", &a[i]);        
    }
    solve();
    return 0;
}