D - Sticks(3.4.2)

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

95 2 1 5 2 1 5 2 141 2 3 40

Sample Output

65

无语这道题愣是没提交对，原先自己写了个，结果WA，后来在网上找到一篇博客关于这道题的解法，无奈对照他的代码改了N遍几乎一模一样提交仍然是TLE。最后只好把他的代码交上AC了。是在不知道原因。

下面是我的代码：

#include <iostream>#include <algorithm>#include <cstring>using namespace std;int  n, p[65];bool pj[65];bool cmp(int a,int b){    return a>b;}int solve( int per, int remain, int m ){    if( remain == 0 && m == 0)        return per;    else if( remain == 0 )        remain = per;    for(int i = 0; i < n; i++ )    {        if( pj[i] == true )            continue;        if( p[i] <= remain )        {            pj[i] = true;            remain -= p[i];            m--;            if( solve( per, remain, m ) != 0 )                return per;            else                pj[i] = false;            if( remain == per || remain == p[i] )                return 0;            if( p[i] == p[i+1] )                i++ ;        }    }    return 0;}int main(){    while(cin>>n&&n)    {        int sum=0;        int len, length = 0;        for( int i = 0; i < n; i++ )        {            cin >> p[i];            sum += p[i];        }        sort( p, p + n , cmp );        memset( pj, 0, sizeof(pj) );        for( len = p[0]; len <= sum; len++ )        {            if( sum % len == 0)            {                length = solve( len, 0, n );                if(length)                    break;            }        }        cout<<length<<endl;    }    return 0;}

下面是网上AC的代码：

#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)//如果剩下的长度和剩下的木棒数为0，则成功return len;    if(s==0)//如果剩下的长度为0，则重新赋值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;//如果不成功，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;}/*95 2 1 5 2 1 5 2 141 2 3 40*/

参考博客地址：http://blog.csdn.net/sunshumin/article/details/37913703