【Ivan and Powers of Two】【CodeForces - 305C】（思维）（Set 应用)

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概述

题目：

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

Ivan wrote out integers 2a1, 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 ≤ 105). The second input line contains nspace-separated integers a1, a2, ..., an (0 ≤ ai ≤ 2·109). It is guaranteed that a1 ≤ a2 ≤ ... ≤ an.

Output

Print a single integer — the answer to the problem.

Examples

Input

4
0 1 1 1

Output

Input

1
3

Output

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.

解题报告：首先咱们需要的数学储备知识，(2^a+2^a)=2^(a+1) ,2^1+2^2……+2^n=2^(n+1)-1;接下来咱们可以开始了，所求的是满足题意的：将所有的数组元素进行2的幂数操作，然后求和，问再添加几位数可以使和满足2^v-1，因为题目没有要求是否最小添加，所以利用set，将数组元素相同的进行合并操作，然后找出最大的数字，之前的用数补齐，那么所补数的数目就是输出的数值。

ac代码：

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
using namespace std;
typedef long long ll;

set<int > ss;
int main()
{
	int n,a;
	while(scanf("%d",&n)!=EOF)
	{
		ss.clear();
		int maxx=0;
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a);
			while(ss.count(a))
			{
				ss.erase(a);
				a++;
			}
			ss.insert(a);
			maxx=max(maxx,a); 
		}
		printf("%dn",maxx-ss.size()+1);
	}
	return 0;
}