SGU

186. The Chain

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

Smith has N chains. Each chain is the sequence of successively connected links. The length of each chain is known: the first chain contains L1 links, the second - L2, ..., the last one - LN. 
He can make a following series of actions in a minute: 
1. to unchain one link 
2. to remove or to put into the unchained link some other links of any chain 
3. to chain the link 
Your task is to determine the minimum time which will take the smith to connect all the chains in one line, i.e. the chain will look like a chain made up of successively connected links.

Input

The first line contains natural number N<=100. The second line contains L1, L2, ..., LN (1<=Li<=100, for all i = 1..N).

Output

Output the only integer number - the solution to the problem.

Sample test(s)

Input

2 3 4

Output

1

[submit]

[forum]

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Author:	Michael R. Mirzayanov
Resource:	ACM International Collegiate Programming Contest 2003-2004  North-Eastern European Region, Southern Subregion
Date:	2003 October, 9

思路：好无耻的英语。说的这么拐弯抹角真的好吗？题意应该是这种。先从一个chain上拆下一个link。然后能够连接两个chain，仅仅到全部chain都连起来为止，这里算一分钟。所以先从短链開始拆更省时，贪心....

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int a[105];

int main() {
	int n;
	while(scanf("%d", &n) != EOF) {
		for(int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
		}
		
		sort(a, a+n);
		
		int i = 0, j = n-1, ans = 0;
		while(i < j) {
			a[i]--;
			j--;
			if(a[i] == 0) i++;
			ans++;
		}
		
		printf("%d
", ans);
	}
	return 0;
}