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  • Educational Codeforces Round 42 (Rated for Div. 2) D. Merge Equals

    题目链接

    D. Merge Equals

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes

    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 kk integers — the elements of the array after all the performed operations.

    Examples

    input

    7
    3 4 1 2 2 1 1

    output

    4
    3 8 2 1

    input

    5
    1 1 3 1 1

    output

    2
    3 4

    input

    5
    10 40 20 50 30

    output

    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.

    AC代码:

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int maxn=3e5+10;
    map<ll,int>mp1,mp2;
    ll ans[maxn],n,x,cnt;
    int main()
    {
        int n;
        cin>>n;
        for(int i=0; i<n; i++)
        {
            cin>>x;
            if(mp1[x]==0)
            {
                mp1[x]++;
                mp2[x]=i;
                ans[i]=x;
            }
            else
            {
                while(mp1[x]!=0)
                {
                    mp1[x]--;
                    ans[mp2[x]]=-1;
                    x*=2;
                    ans[i]=x;
                }
                mp1[x]++;
                mp2[x]=i;
            }
        }
        for(int i=0; i<n; i++)
            if(ans[i]!=-1)
                cnt++;
        cout<<cnt<<endl;
        for(int i=0; i<n; i++)
            if(ans[i]!=-1)
                cout<<ans[i]<<' ';
        cout<<endl;
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/weimeiyuer/p/8831543.html
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