首页 > 代码库 > Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) E. Prairie Partition 二分+贪心
Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) E. Prairie Partition 二分+贪心
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:
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!
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, 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.
8
1 1 2 2 3 4 5 8
2
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_pairtypedef 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;}/*51 2 3 4 5*/
Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) E. Prairie Partition 二分+贪心