zoukankan      html  css  js  c++  java
  • ACM POJ 2356

    Find a multiple

    Time Limit: 1000MS
    Memory Limit: 65536KB
    64bit IO Format: %I64d & %I64u

    [Submit]   [Go Back]   [Status

    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

    Ural Collegiate Programming Contest 1999

    抽屉原理,开始高明白,安叔一点拨就清楚了,Orz,和mod后相同的两个数就可以

    给代码了

    #include <cstdio>
    #include <cstring>
    const int N = 10001;
    int a[N],sum[N],visit[N] ;
     
    int n;
     
    void print(int i,int j)
    {
        printf("%d
    ",j-i+1);
        int k;
        for(k=i; k<=j; k++)
        {
            printf("%d
    ",a[k]);
        }
        return ;
    }
     
    int main()
    {
     
        scanf("%d",&n);
        sum[0] = 0;
        memset(visit,0,sizeof(visit));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            sum[i] = (a[i] + sum[i-1]) % n;
        }
    //    for(int i = 1; i <=n; i++)
    //    {
    //        printf("%d
    ",sum[i]);
    //    }
        for(int i = 1; i<=n; i++)
        {
            if(sum[i] == 0)
            {
                print(1,i);
                return 0;
            }
            else
            {
                if(visit[sum[i]]!= 0)
                {
                    print((visit[sum[i]]+1),i);
                    return 0;
                }
                else
                {
                    visit[sum[i]]=i;
    //                printf("%d
    ",visit[i]);
                }
            }
     
        }
    }
  • 相关阅读:
    (一)Java基本数据类型及运算符
    (二)Java控制执行流程
    ArrayList类源码解析——ArrayList动态数组的实现细节(基于JDK8)
    Java的四个标记接口:Serializable、Cloneable、RandomAccess和Remote接口
    Java容器类源码分析之Iterator与ListIterator迭代器(基于JDK8)
    Java容器类源码分析前言之集合框架结构(基于JDK8)
    浅谈虚树
    点分治
    Ze_Min Tree 主席树
    笛卡尔树的妙用
  • 原文地址:https://www.cnblogs.com/wejex/p/3245131.html
Copyright © 2011-2022 走看看