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  • hdu 1455 Sticks

    Sticks
    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

    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
    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    using namespace std;
    int st[65];
    bool vis[65];
    int cnt;
    int n;
    int l;
    bool cmp(int a,int b)
    {
        return a>=b;
    }
    bool dfs(int num,int len)
    {
        if(num==cnt) return true;
        for(int i=0;i<n;i++)
        {
            if(!vis[i])
            {
                if(len+st[i]==l)
                {
                    vis[i]=true;
                    if(dfs(num+1,0)) return true;
                    vis[i]=false;
    
                }
                if(len+st[i]<l)
                {
                    vis[i]=true;
                    if(dfs(num,len+st[i])) return true;
                    vis[i]=false;
                    if(len==0) return 0;//这里剪枝很重要,而且很经典。如果当前为0,即本次搜索失败,则一定会废弃一个static,则当前长度的分法肯定不成立。
                    int last=st[i];
                    while(last==st[i]) i++;
                }
            }
        }
        return false;
    }
    int main()
    {
    
        while((scanf("%d",&n))&&(n!=0))
        {
            for(int i=0; i<=n; i++) vis[i]=false;
            int sum=0;
            cnt=0;
            for(int i=0; i<n; i++)
            {
                scanf("%d",&st[i]);
                sum+=st[i];
            }
            sort(st,st+n,cmp);
            for(int i=st[0];i<=sum;i++)
            {
    
                if(sum%i==0)
                {
                    //cout<<"fuck"<<endl;
                    cnt=sum/i;
                    l=i;
                    if(dfs(0,0))
                    {
                        printf("%d
    ",l);
                        break;
                    }
                }
            }
        }
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5720425.html
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