Ivan and Powers of Two-优快云博客

Ivan has got an array of n non-negative integers a₁, a₂, ..., a_n. Ivan knows that the array is sorted in the non-decreasing order.

Ivan wrote out integers 2^a₁, 2^a₂, ..., 2^a_n on a piece of paper. Now he wonders, what minimum number of integers of form 2^b (b ≥ 0)need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2^v - 1 for some integer v (v ≥ 0).

Help Ivan, find the required quantity of numbers.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵). The second input line contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 2·10⁹). It is guaranteed that a₁ ≤ a₂ ≤ ... ≤ a_n.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input
4
0 1 1 1

output
0

input
1
3

output
3

Note

In the first sample you do not need to add anything, the sum of numbers already equals 2³ - 1 = 7.

In the second sample you need to add numbers 2⁰, 2¹, 2².

思路：这里写中文思路吧，下面用英文下思路是因为四级还没过关，囧，必须要练练，不然完蛋了有木有啊

贪心。

首先要准备构造等比序列吧

1、如果序列里面没有相同的很好办

2、如果非降序序列有相同的怎么办？

必须合并相同的两个数，直到当前的这个数在集合里面唯一。

理由：

（1）、a，a 合并成a+1是因为：2^a+2^a=2*2^a=2^(a+1),合并成一个和再加一个来和a进行组合是一样的花费

（2）、再说2^a已经是大于低位的所有数字之和，所以出现2个a绝对不会考虑把a拿到低位去和其他数拼凑成低位的其他缺的数。也就是说对于当前ai而言：

不用去管低位，低位有什么就是什么，没什么就得补上，ai只需要往上考虑即可

ＡＣＰｒｏｇｒａｍ：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<set>
using namespace std;
//a,a exists,it must be combined cause 
//(1)a,a->a+1  can minimize the number
//(2)2^a has been bigger than the lower bit numbers,it's impossible to be used to make
//the sum of lower bit numbers.
int main()
{
set<int>ss;
int n;
int a;
cin>>n;
int maxn=0;
for(int i=0;i<n;i++){
   cin>>a;
   while(ss.count(a))ss.erase(a),a++;//2^a+2^a=2*2^a=2^(a+1)//change until 'a' are the only one.
   ss.insert(a);        
   maxn=max(a,maxn); 
}
cout<<maxn+1-ss.size()<<endl;
//system("Pause");
return 0;}