poj 1011 Sticks

Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 149254		Accepted: 35443

Description

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 contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.

Output

The output should 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
0

Sample Output

6
5

Source

Central Europe 1995

#include 
#include
#include
using namespace std;
#define nsize 110
int a[100];
bool visited[100];
bool flag = false;
int n;
int len;

bool cmp(int a, int b) {
	return a > b;
}

void dfs(int dep, int now_len, int u) {   // dep为当前已被用过的小棒数，u为当前要处理的小棒。
	if (flag) return;
	if (now_len == 0) {                    //  当前长度为0，寻找下一个当前最长小棒。
		int k = 0;
		while (visited[k]) k++;              //  寻找第一个当前最长小棒。
		visited[k] = true;
		dfs(dep + 1, a[k], k + 1);
		visited[k] = false;
		return;
	}
	if (now_len == len) {                  //  当前长度为len，即又拼凑成了一根原棒。
		if (dep == n) flag = true;        //  完成的标志：所有的n根小棒都有拼到了。
		else dfs(dep, 0, 0);
		return;
	}
	for (int i = u; i < n; i++)
		if (!visited[i] && now_len + a[i] <= len) {
			if (!visited[i - 1] && a[i] == a[i - 1]) continue;      //  不重复搜索：最重要的剪枝。
			visited[i] = true;
			dfs(dep + 1, now_len + a[i], i + 1);
			visited[i] = false;
		}
}

int main() {
	while (cin >> n, n) {
		fill(a, a + n, 0);
		int max = 0, sum = 0;
		flag = false;
		for (int i = 0; i < n; i++) {
			cin >> a[i];
			if (max < a[i]) {
				max = a[i];
			}
			sum += a[i];
		}
		sort(a, a + n, cmp);
		for (len = max; len <= sum; len++) {
			if (sum%len == 0) {
				fill(visited, visited + n, false);
				dfs(0, 0, 0);
				if (flag) {
					cout << len << endl;
					break;
				}
			}
		}
	}
}