二分

首先假设已经求解到的ans代表合理的最大幸福的，我们显然希望找到一个吃巧克力的方式让他比当前ans更大，否则只能更小。符合二分性质。
而且二分的check函数设计直接把吃过的sum每天减半如果不够了再吃一个新的巧克力，看最后会不会超过n即可

最后调用一次check求道合理的最大幸福度的安排方法。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 5e4 + 7;
ll l, r, mid, ans;
ll n, d, a[N], day[N];
bool check(ll x) {
    ll tot = 0, sum = 0;
    for (int i = 1; i <= d; i++) {
        sum = sum >> 1;
        while (sum < x) {
            sum += a[++tot];
            if (tot > n) return false;
            if (x and x == ans) day[tot] = i;
        }
    }
    return true;
}
int main() {
    n = read(); d = read();
    for (int i = 1; i <= n; i++)
        a[i] = read(), r += a[i];
    while (l <= r) {
        mid = (l + r) >> 1;
        if (check(mid)) ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    printf("%lld\n", ans);
    check(ans);
    for (int i = 1; i <= n; i++) {
        if (day[i]) printf("%lld\n", day[i]);
        else printf("%lld\n", d);
    }
    return 0;
}