HDU1455&&POJ1011-Sticksa

Sticks

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:
65536/32768 K (Java/Others)

                                                                                                        Total Submission(s): 9759    Accepted Submission(s): 2900


Problem 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 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
0

 

Sample Output

6
5

 

Source

ACM暑期集训队练习赛（七）

题意： n根木棍，要求还原为原木棍，且原木棍等长，问最短长度是多少。

</pre><pre name="code" class="cpp">#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <set>

using namespace std;

int a[110],n,sum,cnt,visit[110],flag;

int cmp(int x,int y)
{
return x>y;
}

void dfs(int sum,int x,int length,int pos)
{
if(x==cnt)
{
flag=1;return ;
}
for(int i=pos;i<n;i++)
{
if(visit[i]) continue;
if(sum+a[i]<length)
{
visit[i]=1;
dfs(sum+a[i],x,length,i+1);
if(flag) return ;
visit[i]=0;
if(sum==0) return ;
while(a[i]==a[i+1]) i++;
}
else if(sum+a[i]==length)
{
visit[i]=1;
dfs(0,x+1,length,0);
if(flag) return ;
visit[i]=0;
}
}
}

int main()
{
while(~scanf("%d",&n)&&n)
{
sum=0;flag=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
sort(a,a+n,cmp);
memset(visit,0,sizeof visit);
for(int i=a[0];;i++)
{
if(sum%i!=0) continue;
cnt=sum/i;
dfs(0,1,i,0);
if(flag){printf("%d\n",i);break;}
}
}
return 0;
}