You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value 𝑥 that occurs in the array 2 or more times. Take the first two occurrences of 𝑥 in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, 2⋅𝑥).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like [3,4,1,2,2,1,1]. It will be changed in the following way: [3,4,1,2,2,1,1] → [3,4,2,2,2,1] → [3,4,4,2,1] → [3,8,2,1].
If the given array is look like [1,1,3,1,1] it will be changed in the following way: [1,1,3,1,1] → [2,3,1,1] → [2,3,2] → [3,4].
Input The first line contains a single integer 𝑛 (2≤𝑛≤150000) — the number of elements in the array.
The second line contains a sequence from 𝑛 elements 𝑎1,𝑎2,…,𝑎𝑛 (1≤𝑎𝑖≤109) — the elements of the array.
Output In the first line print an integer 𝑘 — the number of elements in the array after all the performed operations. In the second line print 𝑘 integers — the elements of the array after all the performed operations.
Examples inputCopy 7 3 4 1 2 2 1 1 outputCopy 4 3 8 2 1 inputCopy 5 1 1 3 1 1 outputCopy 2 3 4 inputCopy 5 10 40 20 50 30 outputCopy 5 10 40 20 50 30 Note The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change.
题意： 选出最小的那个出现了至少两次的数，把这个数最左边那个删掉，次左边那个乘以2。 求最终的序列
思路： 优先队列+map模拟（貌似写的常数巨大
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include <set> #include <map> #include <queue> using namespace std; typedef long long ll; const int maxn = 3e5 + 7; map<ll,int>mp; ll a[maxn]; map<ll,int>num,vis; priority_queue<int>q[maxn]; struct Node { ll x; bool operator < (const Node& rhs) const{ if(x == rhs.x) return num[x] < num[rhs.x]; return x > rhs.x; } }; priority_queue<Node>Q; int main() { int n;scanf("%d",&n); int cnt = 0; for(int i = 1;i <= n;i++) { scanf("%lld",&a[i]); num[a[i]]++; if(!mp[a[i]]) mp[a[i]] = ++cnt; q[mp[a[i]]].push(-i); } for(int i = 1;i <= n;i++) { if(!vis[mp[a[i]]]) { Node now; now.x = a[i]; if(num[a[i]] >= 2) Q.push(now); vis[mp[a[i]]] = 1; } } while(!Q.empty()) { Node now = Q.top();Q.pop(); ll x = now.x; num[x] -= 2; int pos1 = -q[mp[x]].top();q[mp[x]].pop(); int pos2 = -q[mp[x]].top();q[mp[x]].pop(); if(!mp[x * 2]) mp[x * 2] = ++cnt; q[mp[x * 2]].push(-pos2); num[x * 2]++; a[pos1] = 0; a[pos2] = x * 2; if(num[x] >= 2) Q.push(now); if(num[x * 2] == 2) { Node now; now.x = x * 2; Q.push(now); } } cnt = 0; for(int i = 1;i <= n;i++) { if(a[i]) cnt++; } printf("%d\n",cnt); for(int i = 1;i <= n;i++) { if(a[i]) printf("%lld ",a[i]); } return 0; }