Merge Equals CodeForces - 962D（优先队列）

CF真是锻炼STL用法的好地方

http://codeforces.com/problemset/problem/962/D
用map实现的做法—》https://blog.youkuaiyun.com/qq_40831340/article/details/81913330
D. Merge Equals
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

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题意：给你n个数，将其中的两两相同的值合并成一个新数放回数列中，直到没有可以合并的数，然后按照原有顺序输出新数列
思路：首先存放数列可以用队列，保持原有顺序，但是题上要求按照小的取，于是想到优先队列存结构体，在结构体中把当前数字的位置保存下来，将数字从对队列中两两往出取，最后会只剩一个，单独判断一下就好，最后结果存在结构体数组中，按照位置排序输出就好

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
set<int> s;
struct node{
    ll num;
    ll pos;
    bool operator <(const node& a)const {
        return num==a.num? pos>a.pos:num>a.num;
    }
}no[150005];
priority_queue <node > q;
int aa[150005];
bool cmp(node a,node b){
    return a.pos<b.pos;
}
int main(){
    ll n;
    cin>>n;
    ll a;
    for(int i=1;i<=n;i++){
        cin>>a;
        no[i].num = a;
        no[i].pos = i;
        q.push(no[i]);
    }
    //cout<<s.size()<<endl;
/*  if(s.size()==n){
        for(int i=1;i<=n;i++) cout<<aa[i]<<" ";
        cout<<endl;
    }*/
    ll k = 0;
    node x,y;
    while(!q.empty()){
        if(q.size()==1){//特判队列只剩一个元素的情况
            no[k++] = q.top();

            q.pop();
        }
        else{
            x = q.top(); q.pop();
            y = q.top(); q.pop();
            if(x.num != y.num){
                no[k++] = x;

                q.push(y);
            }
            else{
                y.num = y.num*2;
                q.push(y);
            }
        }
    }
    cout<<k<<endl;
    sort(no,no+k,cmp);
    for(int i=0;i<k;i++){
        cout<<no[i].num<<" ";
    }
    cout<<endl;
    return 0;
}