Bear and Elections CodeForces - 574A（优先队列）

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本文介绍了一个算法问题，主角Limak为了赢得选举必须通过贿赂选民获得更多票数。文章详细解释了如何利用优先队列来实现最小成本的投票作弊策略。

C - Bear and Elections

CodeForces - 574A

Limak is a grizzly bear who desires power and adoration. He wants to win in upcoming elections and rule over the Bearland.

There are n candidates, including Limak. We know how many citizens are going to vote for each candidate. Now i-th candidate would get a_i votes. Limak is candidate number 1. To win in elections, he must get strictly more votes than any other candidate.

Victory is more important than everything else so Limak decided to cheat. He will steal votes from his opponents by bribing some citizens. To bribe a citizen, Limak must give him or her one candy - citizens are bears and bears like candies. Limak doesn't have many candies and wonders - how many citizens does he have to bribe?

Input

The first line contains single integer n (2 ≤ n ≤ 100) - number of candidates.

The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000) - number of votes for each candidate. Limak is candidate number 1.

Note that after bribing number of votes for some candidate might be zero or might be greater than 1000.

Output

Print the minimum number of citizens Limak must bribe to have strictly more votes than any other candidate.

Example

Input

5
5 1 11 2 8

Output

Input

4
1 8 8 8

Output

Input

2
7 6

Output

Note

In the first sample Limak has 5 votes. One of the ways to achieve victory is to bribe 4 citizens who want to vote for the third candidate. Then numbers of votes would be 9, 1, 7, 2, 8 (Limak would have 9 votes). Alternatively, Limak could steal only 3 votes from the third candidate and 1 vote from the second candidate to get situation 9, 0, 8, 2, 8.

In the second sample Limak will steal 2 votes from each candidate. Situation will be 7, 6, 6, 6.

In the third sample Limak is a winner without bribing any citizen.

思路：每次让人物票数和整个候选人中最大的比较，如果已经大于最大票数了，就完成了，否则，最大票数减1，人物票数加1，然后在重新去选择出最大的票数来，不断循环直到人物票数大于了最大票数，而这个过程使用优先队列可以非常轻松的完成 code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int main(){
    priority_queue<int>q;//从小到大排列，所以是最大堆，先出来的就是最大的
    int n,ans;
    scanf("%d%d",&n,&ans);//ans相当于a[1]
    int i,a;
    for(i = 1; i < n; i++){
        scanf("%d",&a);
        q.push(a);//剩下的全部入队
    }
    int times = 0;//times记录贿赂人的个数
    while(ans<=q.top()){//只要票数小于等于最大的票数
        ans++;//a[1]票数加1
        times++;//人数加1
        a = q.top();
        q.pop();//队首的出去
        q.push(a-1);//减1再进去
    }
    printf("%d\n",times);
    return 0;
}