Code Forces 587A - Duff and Weight Lifting（貪心）

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.

Example

Input

5
1 1 2 3 3

Output

2

Input

4
0 1 2 3

Output

4

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Solution
題目大意：達夫練習舉重，給他n個啞鈴，他每次能舉起重量滿足2a1+2a2+...+2an=2x .的啞鈴。求他需要多少次才能將啞鈴舉完。第一行輸入n表示啞鈴總數，第二行輸入的表示2的係數，數量不超過n。
解題思路：a1、a2…ak如果兩兩不相同，則上面等式不成立，達夫需要n次。由2a+2a=2a+1 ,可以知道兩個小一號的啞鈴可以合併成一個大一號的啞鈴。
可以定義一個數組用來存放2係數的和.
比如輸入 1 1 2 3 3
則cnt[1]=2,cnt[2]=1,cnt[3]=2，2個1可以合併成一個2,則合併後cnt[1]=0,cnt[2]=0,cnt[3]=1,cnt[4]=1。

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Code

#include <stdio.h>
#include <iostream>
#include <algorithm>
#define maxn 1000100
using namespace std;

int cnt[maxn];
int main(int argc, char const *argv[])
{
    int n,a;
    while (~scanf("%d", &n))
    {
        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a);
            cnt[a]++;
        }
        for (int i = 0; i < maxn; i++)
        {
            cnt[i + 1] += cnt[i] / 2;
            cnt[i] = cnt[i] % 2;
            if (cnt[i]) ans++;

        }
        printf("%d\n", ans);
    }
    return 0;
}