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

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

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+, M = 1e3+, mod = 1e9+,inf = 2e9;

LL H[],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 = ; i <= ; ++i) {

        if(x < H[i])

            break;

    }

    i--;

    for(int j = ; j <= i; ++j) {

        cnt[j]--;

        if(cnt[j] < ) No = ;

        return ;

    }

}

int cango(LL x) {

    if(x == ) return ;

    int ok = ;

    for(int i = ; i <= ; ++i) {

        if(H[i] <= x) {

            cnt[i]--;

            if(cnt[i] < ) {

                ok = ;

            }

        }

    }

    if(ok) {

       for(int i = ; i <= ; ++i)

            if(H[i] <= x) cnt[i]++;

        return ;

    }

    else return ;

}

int can(LL now) {

    for(int i = G.size()-; i >= ; --i) {

        int ok = ;

        for(int j = ; j <= ; ++j) {

            if(G[i] <= H[j] && sum[j-]) {

                sum[j-]--;

                P[++cnts] = MP(j-,G[i]);

                ok = ;

                G.pop_back();

                break;

            }

        }

        if(!ok) return ;

    }

    return ;

}

int allcan(int x) {

    int j = x+,i = ;

    int ok;

    while(j <= cnts && i < G.size()) {

        if(P[j].second  != ) j++;

        else if(H[P[j].first+] < G[i]) j++;

        else i++,j++;

    }

    if(i == G.size()) {

        return ;

    }

    else return ;

}

int check(int x) {

    x = cnts - x;

    if(x > cnts) return ;

    if(x == ) return ;

    G.clear();

    for(int i = ; i <= x; i++) {

        for(int j = ; 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 ;

    }

    else return ;

}

int main() {

    H[] = ;

    for(int i = ; i <= ; ++i)H[i] = H[i-]*2LL;

    scanf("%d",&n);

    for(int i = ; i <= n; ++i) {

        scanf("%I64d",&a[i]);

        int ok = ;

        for(int j = ; j <= ; ++j) {

            if(a[i] == H[j]) {

                ok = ;

                cnt[j]++;

                break;

            }

        }

        if(!ok) G.push_back(a[i]);

    }

    int now = ;

    for(int i = ; i >=; --i) {

        while(cnt[i]) {

            if(cango(H[i])) {

             now++;

             sum[i]++;

            }

            else break;

        }

    }

    for(int i = ; i <= ; ++i)

        for(int j = ; j <= cnt[i]; ++j) G.push_back(H[i]);

    int l= ,r,ans = -,tmpr;

    if(can(now)) r = now;

    else r = -;

    tmpr = r;

    for(int i = ; i <= ; ++i) {

        for(int j = ; j <= sum[i]; ++j) {

            P[++cnts] = MP(i,);

        }

    }

    sort(P+,P+cnts+);

    while(l <= r) {

        int md = (l + r) >> ;

        if(check(md)) {

            ans = md;

            r = md-;

        }

        else l = md+;

    }

    //cout<<ans<<endl;

    if(tmpr == -) puts("-1");

    else {

        for(int i = ans; i <= tmpr; ++i) cout<<i<<" ";

        cout<<endl;

    }

    return ;

}

/*

5

1 2 3 4 5

*/

巴特西

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

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