George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.
Input
The input file contains blocks of 2 lines. The first line contains the number of sticks parts after cutting. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.
Output
The output file contains the smallest possible length of original sticks, one per line.
Sample Input
9
5 2 1 5 2 1 5 2 1
4
1 2 3 4
Sample Output
6
5
先从大到小排序，在枚举长度，从最长的棒子开始到总和的一半，负责就是总和，长度一定是总和的约数，
在深搜过程中，如果当前木棒和前一个木棒的长度是一样的，但是前一个木棒没有被选上，那么这个木棒也一定不会被选上。
在深搜过程中，如果当前是在拼一根新木棒的第一截，但如果把可用的最长的一根木棒用上后不能拼成功的话，那么就不用再试后面的木棒了，肯定是前面拼的过程出了问题。
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
using namespace std;
int i, j, n, a[100], sum, c[100], tot, t, f;
void work(int x, int y, int z, int q)
{
int k = 1;
if (z == sum / x) { f = 1; return; }
for (int i = q; i < tot;i++)
if (c[i] && y + a[i] <= x)
{
c[i]--;
if (k&&y == 0) k = 0;
if (y + a[i] < x) work(x, y + a[i], z, i);
else {
work(x, 0, z + 1, 0);
if (f) return;
c[i]++;
break;
}
c[i]++;
if (!k) break;
}
}
int main() {
while (cin >> n, n)
{
for (i = 1, sum = 0, tot = 0; i <= n; i++)
{
cin >> t; sum += t;
for (j = 0; j<tot; j++) if (t == a[j]) break;
if (j == tot) {c[tot] = 0; a[tot++] = t;}
c[j]++;
}
for (i = 0; i<tot;i++)
for (j = i + 1; j<tot;j++)
if (a[i]<a[j]) { swap(a[i], a[j]); swap(c[i], c[j]); }
for (f = 0, i = n; i > 1; i--)
if (sum%i == 0 && sum / i >= a[0]) { work(sum / i, 0, 0, 0); if (f) break; }
end:cout << sum / i<< endl;
}
return 0;
}