Ivan and Powers of Two

 Ivan and Powers of Twotime limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan has got an array of n non-negative integers a_{1, a2, ..., an.
Ivan knows that the array is sorted in the non-decreasing order.}

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

Help Ivan, find the required quantity of numbers.

Input

The first line contains integer n (1 ≤ n ≤ 10^{5).
The second input line contains n space-separated integers a_{1, a2, ..., an (0 ≤ ai ≤ 2·109).
It is guaranteed that a1 ≤ a2 ≤ ... ≤ an.}}

Output

Print a single integer — the answer to the problem.

Sample test(s)input

40 1 1 1

output

0

input

13

output

3

Note

In the first sample you do not need to add anything, the sum of numbers already equals 2^{3 - 1 = 7.}

In the second sample you need to add numbers 2^{0, 21, 22.}

思路：這裡寫中文思路吧，下面用英文下思路是因為四級還沒過關，囧，必須要練練，不然完蛋了有木有啊

貪心。

首先要準備構造等比序列吧

1、如果序列裡面沒有相同的很好辦

2、如果非降序序列有相同的怎麼辦？

必須合并相同的兩個數，直到當前的這個數在集合裡面唯一。

理由：

（1）、a，a 合并成a+1是因為：2^a+2^a=2*2^a=2^(a+1),合并成一個和再加一個來和a進行組合是一樣的花費

（2）、再說2^a已經是大於低位的所有數字之和，所以出現2個a絕對不會考慮把a拿到低位去和其他數拼湊成低位的其他缺的數。也就是說對於當前ai而言：

不用去管低位，低位有什麼就是什麼，沒什麼就得補上，ai只需要往上考慮即可

ＡＣ Ｐｒｏｇｒａｍ：

#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;}