AtCoder Regular Contest 101 C

水题  ；

最优解最多只会回头一次  所以依次判断就行了

链接 ：https://atcoder.jp/contests/arc101/tasks/arc101_a

#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define io std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
const double pi=acos(-1);
const ll P = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n){ll r=1%P;for (a%=P; n; a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
const double eps=1e-5;
ll lcm(ll a,ll b){return a*b/gcd(a,b);};



const int maxn = 1e5+10;
int sum1[maxn],sum2[maxn];
int main(){
    int n,k;
    cin>>n>>k;
    std::vector<int> ans1,ans2;
    for(int i=1;i<=n;i++)
      {
            int x;
            cin>>x;
            if(x<0)ans1.push_back(-x);
            else if(x>0) ans2.push_back(x);
            else k--;
      }
      reverse(ans1.begin(),ans1.end());
      for(int i=0;i<ans2.size();i++)
      { 
        sum1[i+1]=ans2[i];
      }
      for(int i=0;i<ans1.size();i++)
      {
        sum2[i+1]=ans1[i];
      }
      int _min=1e9+10;
      if(k==0)
        _min=0;
      for(int i=1;i<=ans2.size();i++){
        if(k-i>ans1.size())
          continue;
        _min=min(sum1[i]+sum2[k-i]+min(sum1[i],sum2[k-i]),_min);
      }
      for(int i=1;i<=ans1.size();i++)
      {if(k-i>ans2.size())
          continue;
        _min=min(sum2[i]+sum1[k-i]+min(sum2[i],sum1[k-i]),_min);
      }
      cout<<_min<<endl;
    return 0;
}