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
81 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
*/