codeforces Codeforces Round #318 div2 A. Bear and Elections 【优先队列】

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.

Sample test(s)

Input

5
5 1 11 2 8

Output

Input

4
1 8 8 8

Output

Input

2
7 6

Output

0

题目描述：给你n个人，第二行有n个数，每个数代表第i个人的选票数量。现在第1号人想要赢得选举。所以在选票数量上1号候选人的选票数量
必须严格大于其他候选人的选票数。如果1号候选人的选票不能让他被选举上，他就会去投其他人的。直到他的票数最多。问：他最少需要偷多少张？
（偷的选票可以来自其他的任何人）

将2--n号人的选票数加入到一个大数优先的优先队列。只要1号候选人的票数少于对首人的，就将队首元素取出队列&&值-1， 1号人的+1.计数器+1，再将这个
元素加入到优先队列中。直到1号人的票数>队首元素。

code：

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>

using namespace std;

int main()
{
    int n;
    while(scanf("%d", &n)!=EOF)
    {
        int head;
        priority_queue<int, vector<int>, less<int> >q;
        scanf("%d", &head);
        int cur;
        for(int i=2; i<=n; i++){
            scanf("%d", &cur); q.push(cur);
        }
        int ans=0;
        while(head<=q.top())
        {
            cur=q.top(); q.pop();
            head++; cur--; ans++;
            q.push(cur);
        }
        printf("%d
", ans);
    }
    return 0;
}