POJ 1011 Sticks

最新推荐文章于 2024-04-30 20:38:26 发布

转载最新推荐文章于 2024-04-30 20:38:26 发布 · 55 阅读

本文介绍了一种解决如何从已知的木棒碎片长度推算出原始木棒最小可能长度的问题。通过递归深度优先搜索算法实现，适用于数量不超过64的木棒碎片。

这个问题是做哭= =
之前我们把所有变量定义为全局变量，结果总是提交的时间RT，我无法找到故障。后来改为局部变量ac该= =。

D - Sticks(3.4.2)

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1011

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

题目大意：

任意分割木棒，分割后想把这些碎木棒还原到原来状态，可是他忘记了原先有多少木棒以及那些木棒原来有多长。请计算原先木棒的最小可能长度。

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int n,num[69];
int pd[69];
int cmp(int a,int b)
{
	return a>b;
}
int dfs(int len,int s,int nu)
{
	int i;
    if(s==0&&nu==0)//假设剩下的长度和剩下的木棒数为0。则成功
		return len;
    if(s==0)//假设剩下的长度为0，则又一次赋值
		s=len;
    for(i=0;i<n;i++)
    {
        if(pd[i]==1)
			continue;
        if(s>=num[i])
        {
            pd[i]=1;
            if(dfs(len,s-num[i],nu-1))
				return  len;	
			pd[i]=0;//假设不成功，则当前棒不使用
            if(num[i]==s||s==len)
				break;
            while(num[i]==num[i+1])
				i++;			
        }
    }
	return 0;
}
int main()
{
    while(cin>>n&&n)
    {
		int i;
		int sum;
        sum=0;
		int len,k;
		for(i=0;i<n;i++)
		{
			cin>>num[i];
			sum=sum+num[i];
		}
		sort(num,num+n,cmp);
		for(len=num[0];len<=sum;len++)
		{
			memset(pd,0,sizeof(pd));
			if(sum%len==0)
			{
				k=dfs(len,0,n);
				if(k)
					break;
			}
		}
		cout<<k<<endl;
    }
    return 0;
}
/*
9
5 2 1 5 2 1 5 2 1
4
1 2 3 4
0
*/