*搜索（剪剪剪枝）POJ - 1011 Sticks

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

代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define maxn 70
#define INF 0x3f3f3f3f
using namespace std;
bool vis[maxn];
int n;
int a[maxn];
int len;
bool cmp(int x,int y)
{
    return x>y;
}
bool dfs(int ans,int x)
{
    if(ans==0)
    {
        return 1; //所有的都被分配
    }
    if(x==0) x=len;
    for(int i=0; i<n; i++)
    {
        if(vis[i]||a[i]>x) continue;  //a[i]被用或者a[i]長度大於所需長度
        vis[i]=1;
        if(dfs(ans-1,x-a[i])) return true; //存放成功返回上一層
        vis[i]=0;
        if(x==len||x==a[i]) return false; //長度足夠但不能被存滿而返回到原是長度或a[i]
        while(a[i]==a[i+1])i++; //重複元素不搜索
    }
    return false;
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        int sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n,cmp);
        int i;
        for(i=a[0]; i<=sum/2; i++)
        {
            if(sum%i) continue;
            memset(vis,0,sizeof(vis));
            len=i;
            if(dfs(n,i))
            {
                printf("%d\n",i);
                break;
            }
        }
        if(i>sum/2) printf("%d\n",sum);
    }
    return 0;

}

怪不得網上代碼都差不多，自能說要剪枝的地方太多，少了那個又不行，錯了N次，照個學長的代碼看着改都不行，我會常來看你的