[USACO 2010 Feb S]Chocolate Eating

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whiteawll 2022-08-16 14:40:49 博主文章分类：练习题目 ©著作权

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// Problem: [USACO 2010 Feb S]Chocolate Eating
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/22353/E
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-04-19 15:51:21
// 
// Powered by CP Editor (https://)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
    ll ans = 0;
    while (y) {
      if (y & 1)
         ans = (ans + x) %mod;
      y >>= 1;
      x = (x + x) %mod;
   }
   return ans;
}
inline ll qmi (ll a, ll b) {
   ll ans = 1;
   while (b) {
      if (b & 1) ans = ans * a;
      a = a * a;
      b >>= 1;
   }
   return ans;
}
inline int read () {
   int x = 0, f = 0;
   char ch = getchar();
   while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
   while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
   return f?-x:x;
}
template<typename T> void print(T x) {
   if (x < 0) putchar('-'), x = -x;
   if (x >= 10) print(x/10);
   putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
   return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
   return (a + b) %mod;
}
// inline ll inv (ll a) {
   // return qmi(a, mod - 2);
// }
#define int long long
int n, d;
int a[100001];
int ans[100001], s;
bool check (int tar) {
   int sum = 0;
   int k = 0;
   for (int i = 1; i<= d; i ++) {
         sum /= 2;
         while (sum < tar) {
            k ++;
            sum += a[k];
            if (k > n) return 0;
            if (tar && tar == s) {
               ans[k] = i;
            }
         }
   }
   return 1;
}
void solve() {
   cin >> n >> d;
    int l = 0, r = 0;
    for (int i = 1; i<= n; i ++) {
        cin >> a[i];
        r+= a[i];
    }
   
    
     while (l < r) {
         int mid = l + r+ 1>>1;
         if (check(mid)) {
            s = mid;
            l = mid;
         }
         else r = mid - 1;
     }
     cout << s << endl;
     check(s);
     for (int i = 1; i<= n; i ++) {
         if (ans[i] == 0)
                ans[i] = d;
      cout << ans[i] << endl;
     }
}
signed main () {
   // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}