整点巧克力(二分)

整点巧克力

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

const int N = 5e4 + 10;

long long a[N], sum;
int n, d;
int s[N];

bool check(long long mid){
    long long tmp = 0;
    int cnt = 0;
    for(int i = 1; i <= d; ++i){
        while(tmp < mid){
            tmp += a[++cnt];
            if(cnt > n) return false;
            s[cnt] = i;
        }
        tmp /= 2;
    }
    for(int i = cnt + 1; i <= n; ++i) s[i] = d;
    
    return true;
}
signed main(){
    cin >> n >> d;
    for(int i = 1; i <= n; ++i){
        cin >> a[i]; sum += a[i];
    }
    long long l = 0, r = sum;
    //二分快乐值
    while(l < r){
        long long mid = (l + r + 1) >> 1;
        if(check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << l << endl;
    check(l);
    for(int i = 1; i <= n; ++i) cout << s[i] << endl;
    
    return 0;
}