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334B.cpp
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334B.cpp
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/*
* J1K7_7
*/
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <cstring>
#include <list>
#include <map>
#include <unordered_map>
#include <iomanip>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <limits>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<long long> vll;
#define l(x) (x << 1) + 1
#define r(x) (x << 1) + 2
#define mid(l, r) ((l + r) >> 1)
#define mp make_pair
#define pb push_back
#define all(a) a.begin(),a.end()
#define debug(x) {cerr <<#x<<" = " <<x<<"\n"; }
#define debug2(x, y) {cerr <<#x<<" = " <<x<<", "<<#y <<" = " <<y <<"\n";}
#define ss second
#define ff first
#define m0(x) memset(x,0,sizeof(x))
#define snuke(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
template<class T>
inline bool ispow2(T x){return (x!=0 && (x&(x-1))==0);} //0 or 1
template<class T> inline T powmod(T a,T b,T mod) {ll res = 1; while(b){if(b&1) res = (res*a)%mod;a = (a*a)%mod;b >>= 1;}return res;}
template<class T> inline T gcd(T a,T b){ll t;while(b){a=a%b;t=a;a=b;b=t;}return a;}
template<class T> inline T lcm(T a,T b){return a/gcd(a,b)*b;}
inline int nextint(){ int x; scanf("%d",&x); return x; }
inline ll nextll(){ ll x; scanf("%lld",&x); return x; }
const int mod=1e9+7;
const ll mx_ll = numeric_limits<ll> :: max();
const int mx_int = numeric_limits<int> :: max();
const long double PI = (long double)(3.1415926535897932384626433832795);
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
int n, k; cin >> n >> k;
vector<ll> s(n);
for(int i = 0;i < n ;i++) cin >> s[i];
ll ans = *max_element(all(s));
if( n <= k )
{
cout << ans;
return 0;
}
int i = 0;
//n-k box pair and 2k-n single's
// n -(2k-n) = 2(n-k) cowbells
while(i < n-k)
{
ans = max( ans , s[i] + s[2*(n-k)-1-i]);
i++;
}
cout << ans ;
return 0;
}