acm 1011 Sticks

本文介绍了一个编程问题——如何通过已知切割后的棍子长度来计算原始棍子的最短可能长度。文章提供了一段C++代码示例,该程序能够处理用户输入的数据,并通过算法找出最小的原始棍长。

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Sticks
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 113105 Accepted: 25961

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

#include<iostream>
#include <stdio.h> 
#include <stdlib.h>
using namespace std;
int num[64];
bool mg[64];
int n;
int cmp(const void *i, const void *j)
{
	return *(int *)j - *(int *)i;
}
bool find(int m,int k,int len)
{
     if(m==0 && k==0)
         return true;
     if(m==0)
        m=len;
     for(int i=0;i<n;i++)    
     {
          if(!mg[i] && num[i]<=m)
          {
              mg[i]=true;
              if(find(m-num[i],k-1,len))
                 return true;
              mg[i]=false;
              if (m == num[i] || m == len)
               return false;            
          }        
     }
     return false;
}
int main()
{
    int i,max,sum;
    while(true)
    {
       scanf("%d",&n);
       if(n==0)
         break;
       max=0;
       sum=0;
       for(i=0;i<n;i++)
       {
            scanf("%d",&num[i]);
            sum=sum+num[i];  
            mg[i]=false;            
       }
       qsort(num, n, sizeof(int), cmp);
       for(int j=num[0];j<=sum;j++)
          if((sum%j==0) && find(j,n,j))
           {
             cout<<j<<endl;
             break;
           }
    }
    system("pause");
    return 0;
}


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