Educational Codeforces Round 42 (Rated for Div. 2) D.Merge Equals(data structures,STL)

D. Merge Equals
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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 xx that occurs in the array 22 or more times. Take the first two occurrences of xx 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, 2x2⋅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][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][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][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][1,1,3,1,1] → [2,3,1,1] → [2,3,2] → [3,4].

Input

The first line contains a single integer nn (2n1500002≤n≤150000) — the number of elements in the array.

The second line contains a sequence from nn elements a1,a2,,ana1,a2,…,an (1ai1091≤ai≤109) — the elements of the array.

Output

In the first line print an integer kk — the number of elements in the array after all the performed operations. In the second line print kkintegers — the elements of the array after all the performed operations.

Examples
input
Copy
7
3 4 1 2 2 1 1
output
Copy
4
3 8 2 1 
input
Copy
5
1 1 3 1 1
output
Copy
2
3 4 
input
Copy
5
10 40 20 50 30
output
Copy
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.

不得不惊叹map,set等STL的神奇用处,但是这种地址迭代器的操作我还是生疏的很,在省赛到来之前需要恶补一波了~~

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<time.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<algorithm>
#define INF 0x3f3f3f3f
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define FAST_IO ios::sync_with_stdio(false)
#define mem(a,b) memset(a,b,sizeof(a))
const double PI = acos(-1.0);
const double eps = 1e-6;
typedef unsigned long long ll;
using namespace std;

const int N=200005;
map< ll,set<int> >m;
ll a[N];
int main()
{
    int n;
    scanf("%d",&n);

    int all=n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        m[a[i]].insert(i);
    }

    for(map< ll,set<int> >::iterator i=m.begin();i!=m.end();i++)
    {
        while(i->second.size()>1)
        {
            set<int>::iterator op1=i->second.begin(),op2;
            op2=op1;op2++;
            all--;
            int t1=*op1,t2=*op2;
            a[t1]=-1;
            a[t2]*=2;
            m[a[t2]].insert(t2);
            i->second.erase(op1);
            i->second.erase(op2);
        }
    }
    cout<<all<<endl;
    for(int i=1;i<=n;i++)
    {
        if(a[i]!=-1)
            cout<<a[i]<<" ";
    }
    return 0;
}

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转载自blog.csdn.net/sudu6666/article/details/80055202