Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) E. Prairie Partition 二分+贪心

Posted 2020-09-17 Meek

E. Prairie Partition

It can be shown that any positive integer x can be uniquely represented as x = 1 + 2 + 4 + ... + 2k - 1 + r, where k and r are integers, k ≥ 0, 0 < r ≤ 2k. Let‘s call that representation prairie partition of x.

For example, the prairie partitions of 12, 17, 7 and 1 are:

12 = 1 + 2 + 4 + 5,

17 = 1 + 2 + 4 + 8 + 2,

7 = 1 + 2 + 4,

1 = 1.

Alice took a sequence of positive integers (possibly with repeating elements), replaced every element with the sequence of summands in its prairie partition, arranged the resulting numbers in non-decreasing order and gave them to Borys. Now Borys wonders how many elements Alice‘s original sequence could contain. Find all possible options!

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of numbers given from Alice to Borys.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1012; a1 ≤ a2 ≤ ... ≤ an) — the numbers given from Alice to Borys.

Output

Output, in increasing order, all possible values of m such that there exists a sequence of positive integers of length m such that if you replace every element with the summands in its prairie partition and arrange the resulting numbers in non-decreasing order, you will get the sequence given in the input.

If there are no such values of m, output a single integer -1.

Examples

input

8
1 1 2 2 3 4 5 8

output

2 

Note

In the first example, Alice could get the input sequence from [6, 20] as the original sequence.

In the second example, Alice‘s original sequence could be either [4, 5] or [3, 3, 3].

 题意：

　　每个数都可以表示成2的连续次方和加上一个r

　　例如：12 = 1 + 2 + 4 + 5,

　　17 = 1 + 2 + 4 + 8 + 2,

　　现在给你这些数，让你反过来组成12，17，但是是有不同方案的

　　看看样列就懂了，问你方案的长度种类

题解：

　　　将所有连续的2^x，处理出来，假设有now个序列

　　 最后剩下的数，我们必须将其放到上面now的尾端，但是我们优先放与当前值最接近的序列尾端，以防大一些的数仍然有位置可以放

　　 处理出满足条件最多序列数

　　二分最少的能满足条件的序列数，也就是将mid个序列全部插入到上面now-mid个序列尾端，这里贪心选择2^x，x小的

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 1e5+10, M = 1e3+20, mod = 1e9+7,inf = 2e9;

LL H[70],a[N];
int No,cnt[N],n,cnts;
vector<LL > G,ans;
vector<LL > all[N];
int sum[N],sum2[N];
pair<int,LL> P[N];


void go(LL x) {
    int i;
    for(i = 0; i <= 60; ++i) {
        if(x < H[i])
            break;
    }
    i--;
    for(int j = 0; j <= i; ++j) {
        cnt[j]--;
        if(cnt[j] < 0) No = 1;
        return ;
    }
}
int cango(LL x) {
    if(x == 0) return 0;
    int ok = 0;
    for(int i = 0; i <= 60; ++i) {
        if(H[i] <= x) {
            cnt[i]--;
            if(cnt[i] < 0) {
                ok = 1;
            }
        }
    }
    if(ok) {
       for(int i = 0; i <= 60; ++i)
            if(H[i] <= x) cnt[i]++;
        return 0;
    }
    else return 1;
}
int can(LL now) {
    for(int i = G.size()-1; i >= 0; --i) {
        int ok = 0;
        for(int j = 1; j <= 60; ++j) {
            if(G[i] <= H[j] && sum[j-1]) {
                sum[j-1]--;
                P[++cnts] = MP(j-1,G[i]);
                ok = 1;
                G.pop_back();
                break;
            }
        }
        if(!ok) return 0;
    }
    return 1;
}
int allcan(int x) {
    int j = x+1,i = 0;
    int ok;
    while(j <= cnts && i < G.size()) {
        if(P[j].second  != 0) j++;
        else if(H[P[j].first+1] < G[i]) j++;
        else i++,j++;
    }
    if(i == G.size()) {
        return 1;
    }
    else return 0;
}
int check(int x) {
    x = cnts - x;
    if(x > cnts) return 0;
    if(x == 0) return 1;
    G.clear();
    for(int i = 1; i <= x; i++) {
        for(int j = 0; j <= P[i].first; ++j) {
            G.push_back(H[j]);
        }
        if(P[i].second) {
            G.push_back(P[i].second);
        }
    }
    //for(int i = 0; i < G.size(); ++i) cout<<G[i]<<" ";cout<<endl;
    if(allcan(x)) {
        return 1;
    }
    else return 0;
}
int main() {
    H[0] = 1;
    for(int i = 1; i <= 60; ++i)H[i] = H[i-1]*2LL;
    scanf("%d",&n);
    for(int i = 1; i <= n; ++i) {
        scanf("%I64d",&a[i]);
        int ok = 0;
        for(int j = 0; j <= 60; ++j) {
            if(a[i] == H[j]) {
                ok = 1;
                cnt[j]++;
                break;
            }
        }
        if(!ok) G.push_back(a[i]);
    }
    int now = 0;
    for(int i = 60; i >=0; --i) {
        while(cnt[i]) {
            if(cango(H[i])) {
             now++;
             sum[i]++;
            }
            else break;
        }
    }
    for(int i = 0; i <= 60; ++i)
        for(int j = 1; j <= cnt[i]; ++j) G.push_back(H[i]);
    int l= 1,r,ans = -1,tmpr;
    if(can(now)) r = now;
    else r = -1;
    tmpr = r;
    for(int i = 0; i <= 60; ++i) {
        for(int j = 1; j <= sum[i]; ++j) {
            P[++cnts] = MP(i,0);
        }
    }
    sort(P+1,P+cnts+1);
    while(l <= r) {
        int md = (l + r) >> 1;
        if(check(md)) {
            ans = md;
            r = md-1;
        }
        else l = md+1;
    }
    //cout<<ans<<endl;
    if(tmpr == -1) puts("-1");
    else {
        for(int i = ans; i <= tmpr; ++i) cout<<i<<" ";
        cout<<endl;
    }
    return 0;
}
/*
5
1 2 3 4 5
*/