题干：
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
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.
题目大意：（可以练读题）
给你n个数a1,a2....an，分表代表2^a1,2^a2....2^an。现在让你添加一些数（假设num个），使2^a1 + 2^a2 + ... + 2^an + ... + 2^anum 的和为二进制的全1数（题目中叙述是存在一个b，s.t. 数字 = 2^b-1）。求num的最小值
解题报告：
推出来，需要的数恰好就是最终集合合并后的剩下缺少的元素之后（不难证明吧貌似2333），接下来就是怎么维护这个集合的元素了。
这题可以贪心，用pq暴力合并，别用multiset+set、、、会炸的（不是炸在迭代器上而是炸在多次的insert上）、
当然还有更好的做法那就是直接set，然后读入的同时就维护set中的元素就可以了，最后直接输出一个值就是我们需要的那个值了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
set<ll> ss;
int main()
{int n;ll tmp,maxx = -1;cin>>n;for(int i = 1; i<=n; i++) {scanf("%lld",&tmp);while(ss.find(tmp) != ss.end()) {ss.erase(tmp);tmp++;}ss.insert(tmp);maxx = max(maxx,tmp);} printf("%lld\n",maxx + 1 - ss.size());return 0 ;}错误代码：（这样写思路是没错的就是insert的时候需要insert tmp/2 这么多个，所以时间效率还是不高，改成map可能会好一点）（刚开始不是+1，，是+tmp/2，这样是一定错的，因为这是幂次关系啊不是线性的、、）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int n;
ll tmp;
multiset<ll> ms;
set<ll> s;
ll a[MAX];
int main() {cin>>n;for(int i = 1; i<=n; i++) scanf("%lld",a+i),ms.insert(a[i]),s.insert(a[i]);for(set<ll> :: iterator it = s.begin(); it != s.end();) {
// while(ms.count(*it) >= 2) {
// ms.erase(ms.find((*it)));
// ms.erase(ms.find((*it)));
// ms.insert((*it)+1);
// s.insert((*it)+1);
// if(ms.find((*it)) == ms.end()) s.erase((*it));
// }
// ++it;tmp = ms.count((*it));if(tmp % 2 == 1) {s.insert((*it) + 1);ms.erase(*it);ms.insert(*it);ms.insert((*it) + 1);} else {ms.erase((*it));s.insert((*it) + 1);ms.insert((*it) + 1);s.erase(*it);}++it;}multiset<ll> :: iterator mit = ms.end();--mit;
// cout << *mit<< endl;ll ans = (*mit) - s.size() + 1;cout << ans << endl;return 0 ;
}
