zoukankan      html  css  js  c++  java
  • POJ-2356 Find a multiple(DFS,抽屉原理)

    Find a multiple
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 7133 Accepted: 3122 Special Judge
    Description

    The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).
    Input

    The first line of the input contains the single number N. Each of next N lines contains one number from the given set.
    Output

    In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order.

    If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.
    Sample Input

    5
    1
    2
    3
    4
    1
    Sample Output

    2
    2
    3

    dfs:

    #include <iostream>
    #include <string.h>
    #include <math.h>
    #include <stdlib.h>
    #include <algorithm>
    
    using namespace std;
    int a[10005];
    int vis[10005];
    int n;
    int m;
    int dfs(int x,int sum)
    {
        if(sum%n==0&&sum>=n)
        {
            cout<<m<<endl;
            return 1;
        }
        for(int i=x+1;i<=n;i++)
        {
            m++;
            if(dfs(i,sum+a[i]))
            {
                    cout<<a[i]<<endl;
                    return 1;
            }
            m--;
        }
        return 0;
    
    }
    int main()
    {
    
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=1;i<=n;i++)
                scanf("%d",&a[i]);
            m=0;
            if(!dfs(0,0))
                printf("0
    ");
    
        }
         return 0;
    }

    抽屉原理改天补上

  • 相关阅读:
    linux命令
    常用正则表达式总结
    List集合对象根据字段排序
    IO字 节流/字符流 读取/写入文件
    Jquery广告浮动效果小案例
    拿到添加对象的id号方法
    Jquery省市区三级联动案例
    JAVA集合迭代遍历和特性介绍
    Listener监听器使用小案例
    java中用过滤器解决字符编码问题
  • 原文地址:https://www.cnblogs.com/dacc123/p/8228808.html
Copyright © 2011-2022 走看看