zoukankan      html  css  js  c++  java
  • [POJ2356] Find a multiple 鸽巢原理

    Find a multiple
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 8776   Accepted: 3791   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
    

    Source

     
     

     
    题解:
     
    我们可以求出每个数的前缀和,如果有一项mod n等于0,那么直接输出它之前的所有数;
    如果不存在,那么qzh[i]%n的值一定落在[1,n-1]之间,根据鸽巢原理,n个数落在n-1个地方,必定有一个地方重复,即qzh[i] % n = qzh[j] % n;
    所以qzh[i]%n - qzh[j]%n = 0, 即i 到 j 之间的所有数加起来就是n的倍数;
    所以直接暴力判断ok;
     

     
    Code:
    #include <iostream>
    #include <cstdio>
    #include  <map>
    using namespace std;
    
    int n;
    int a[10010];
    int qzh[10010];
    map <int, int> mp;
    
    int main()
    {
        scanf("%d", &n);
        for (register int i = 1 ; i <= n ; i ++) scanf("%d", a + i);
        for (register int i = 1 ; i <= n ; i ++)
        {
            qzh[i] = qzh[i-1] + a[i];
            if (qzh[i] % n == 0) 
            {
                cout << i << endl;
                for (register int j = 1 ; j <= i ; j ++) printf("%d
    ", a[j]);
                return 0;
            }
            if (mp[qzh[i]%n]!= 0) 
            {
                cout << i - mp[qzh[i]%n] << endl;
                for (register int j = mp[qzh[i]%n] + 1 ; j <= i ; j ++)
                    printf("%d
    ", a[j]);
                break;
            }
            mp[qzh[i]%n] = i;
        }
        return 0;
    }
  • 相关阅读:
    Elasticsearch安装中文分词器IK
    Docker安装ElasticSearch
    Docker安装
    Docker安装EOS
    Docker更改容器端口映射
    BoltDB 一个简单的纯 Go key/value 存储
    go语言 robfig/cron包 实现定时 调用
    GitHub 添加 SSH keys
    Android 端外推送到底有多烦?
    Nats的消息通信模型
  • 原文地址:https://www.cnblogs.com/BriMon/p/9251714.html
Copyright © 2011-2022 走看看