zoukankan      html  css  js  c++  java
  • 抽屉问题 Find a multiple POJ

    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

    思路: 打表可知一定有一段区间的解
    前缀和一下 发现 满足的条件为 s[i]%n==s[j]%n (*)
    又因为 0到 n 有 n+1个数
    但是前缀和的答案为 0 到 n-1 有 n 个数
    所以一定有两个前缀和满足(*)的条件
    暴力即可
    code:
    //
    #include<iostream>
    using namespace std;
    int n;
    #define maxnn 20000
    int sum[maxnn];
    int a[maxnn];
    int main()
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            sum[i]=(sum[i-1]+a[i])%n;
        }
        for(int i=1;i<=n;i++)
            for(int j=i;j<=n;j++)
            {
                if(sum[i-1]==sum[j])
                {
                    cout<<j-i+1<<endl;
                    for(int k=i;k<=j;k++)
                    {
                        cout<<a[k]<<endl;
                    }return 0;
                }
            }
        
        
    }


    刀剑映出了战士的心。而我的心,漆黑且残破
  • 相关阅读:
    使用animate()完成修改图片src切换图片的动画效果
    一键分享到各个SNS插件
    $data[$i++]+=2;不等于$data[$i++]=$data[$i++]+2;
    QQ在线客服的使用
    JQuery实时监控文本框字符变化
    迭代器的使用
    泛型的作用
    Eclipse的使用
    关于“类型”字段的处理
    java servlet+mysql全过程(原创)
  • 原文地址:https://www.cnblogs.com/OIEREDSION/p/11308385.html
Copyright © 2011-2022 走看看