Codeforces Round #326 (Div. 2) C. Duff and Weight Lifting 水题

C. Duff and Weight Lifting

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/588/problem/C

Description

Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.

Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.

Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.

Input

The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.

The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.

Output

Print the minimum number of steps in a single line.

Sample Input

5
1 1 2 3 3

Sample Output

2

HINT

题意

给你n个数，然后你可以一次性消去一些数，只要满足了2^a1 + 2^a2 + ... + 2^ak = 2^x这个条件就好

题解：

有一个贪心，我们尽量用大的数去换就好了

于是就可以得到下面这个代码啦

代码：

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<map>
#include<cstring>
using namespace std;

long long a[3000005];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        long long x;
        scanf("%lld",&x);
        a[x]++;
    }
    long long ans =0;
    for(int i = 0;i<=2000005;i++)
    {
        ans+=(a[i]&1);
        a[i+1]+=a[i]/2;
    }
    cout<<ans<<endl;
}