[POJ3370]&[HDU1808]Halloween treats

Description

-Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided.

Your job is to help the children and present a solution.

-Input:The input contains several test cases.

The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a1 , ... , an (1 ≤ ai ≤ 100000 ), where ai represents the number of sweets the children get if they visit neighbour i.

The last test case is followed by two zeros.

-Output:

For each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of ai sweets). If there is no solution where each child gets at least one sweet print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.

Solution

1.本题与Find a multiple(题解随笔:http://www.cnblogs.com/COLIN-LIGHTNING/p/8481478.html)基本相同,只需多组数据处理,数据一定要先全部读完再处理;

2.由于n<m,那么至少有两个邻居给的糖果数前缀和关于孩子数同余,取模后相同的前缀和后出现的地址减去前面的地址即可;

3.打一个地址标记,记录模值的第一次出现地址,当第二次出现同一模后前缀和时,使l=第一次出现地址,r=第二次出现地址,区间[l,r]即为所求,注意题目要求输出的是可能的一组地址;

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int i,j,l,r,t[100001]={},a[100001]={},sum[100001]={};
void cul(int n,int m){ //分组处理数据;
l=0;
memset(t,-1,sizeof(t));
memset(a,0,sizeof(a));
memset(sum,0,sizeof(sum));
t[0]=0;
for(i=1;i<=m;++i) scanf("%d",&a[i]);
for(i=1;i<=m;++i){
sum[i]=(sum[i-1]+a[i])%n;
if(t[sum[i]]!=-1){
l=t[sum[i]];
r=i;
break;
}
else t[sum[i]]=i; //计录模值首次出现的地址;
}
for(i=l+1;i<=r;++i)printf("%d ",i); //注意输出地址;
printf("\n");
}
int main(){
int n,m;
scanf("%d%d",&n,&m);
while(n!=0){
cul(n,m);
scanf("%d%d",&n,&m);
}
return 0;
}

有关鸽巢原理可以参考我的博客:http://www.cnblogs.com/COLIN-LIGHTNING/p/8439555.html

[POJ3370]&[HDU1808]Halloween treats 题解(鸽巢原理)的更多相关文章

  1. POJ3370&amp;HDU1808 Halloween treats【鸽巢原理】

    题目链接: id=3370">http://poj.org/problem?id=3370 http://acm.hdu.edu.cn/showproblem.php?pid=1808 ...

  2. POJ 3370 Halloween treats( 鸽巢原理简单题 )

    链接:传送门 题意:万圣节到了,有 c 个小朋友向 n 个住户要糖果,根据以往的经验,第i个住户会给他们a[ i ]颗糖果,但是为了和谐起见,小朋友们决定要来的糖果要能平分,所以他们只会选择一部分住户 ...

  3. poj 3370 Halloween treats(鸽巢原理)

    Description Every year there is the same problem at Halloween: Each neighbour is only willing to giv ...

  4. POJ 3370. Halloween treats 抽屉原理 / 鸽巢原理

    Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798 ...

  5. POJ 3370 Halloween treats 鸽巢原理 解题

    Halloween treats 和POJ2356差点儿相同. 事实上这种数列能够有非常多,也能够有不连续的,只是利用鸽巢原理就是方便找到了连续的数列.并且有这种数列也必然能够找到. #include ...

  6. [HDU1205]吃糖果 题解(鸽巢原理)

    [HDU1205]吃糖果 Description -HOHO,终于从Speakless手上赢走了所有的糖果,是Gardon吃糖果时有个特殊的癖好,就是不喜欢将一样的糖果放在一起吃,喜欢先吃一种,下一次 ...

  7. [POJ2356]Find a multiple 题解(鸽巢原理)

    [POJ2356]Find a multiple Description -The input contains N natural (i.e. positive integer) numbers ( ...

  8. POJ 2356. Find a multiple 抽屉原理 / 鸽巢原理

    Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   ...

  9. poj2356 Find a multiple(抽屉原理|鸽巢原理)

    /* 引用过来的 题意: 给出N个数,问其中是否存在M个数使其满足M个数的和是N的倍数,如果有多组解, 随意输出一组即可.若不存在,输出 0. 题解: 首先必须声明的一点是本题是一定是有解的.原理根据 ...

随机推荐

  1. nginx 配置文件简介

    主配置文件说明(先将注释部分去掉:sed -ri ‘/^#|[[:space:]]+#/d’ /etc/nginx/nginx.conf) (1)全局配置段 1:指明运行worker进程的用户和组 u ...

  2. 结对编程学习fault、error、failure三种状态

    点滴成就 学习时间 新编写代码行数 博客量(篇) 学习知识点 第一周 10小时 0 0 了解软件工程 第二周 10小时 0 1 项目开题 第三周 15小时 0 1 开通博客.开展项目调查 第四周 20 ...

  3. Ubuntu 16.04出现:Problem executing scripts APT::Update::Post-Invoke-Success 'if /usr/bin/test -w /var/

    转自:http://blog.csdn.net/zzq123686/article/details/77454066 出现错误信息: Reading package lists... Done E:  ...

  4. phaser3 微信小游戏若干问题

    纯属个人兴趣, 如有兴趣可共同参与维护. git: https://gitee.com/redw1234567/phaser3_wx image的地方需要修改,代码贴上 var ImageFile = ...

  5. nodejs之Buffer

    Buffer是什么? 简单点理解,buff就是固定长度的uint8array.(es6已实现TypedArray). 由于是固定长度所以没有了splice,concat方法. 由于是固定类型所以没有了 ...

  6. utuntu 安装python3.5

    如果想要升级Utuntu系统中的python版本,请不要卸载原先的版本. 桌面环境中的需要依赖于python相关,卸载之后会出现意想不到问题. (1)sudo add-apt-repository p ...

  7. 【服务器_Tomcat】Tomcat的Server Options选项

    一.配置 默认前两个是没有勾选的,应该勾选上: 在Cotext节点中有一个reloadable='true'属性,将它改为false,可以在修改java文件后不用重启服务器即可生效,但是不包括新建ja ...

  8. [九]SpringBoot 之 定时任务

    代码: package me.shijunjie.config; import org.springframework.context.annotation.Configuration; import ...

  9. BZOJ4881 线段游戏(二分图+树状数组/动态规划+线段树)

    相当于将线段划分成两个集合使集合内线段不相交,并且可以发现线段相交等价于逆序对.也即要将原序列划分成两个单增序列.由dilworth定理,如果存在长度>=3的单减子序列,无解,可以先判掉. 这个 ...

  10. Windows与VMware中的CentOS系统互通访问

    [步骤01]设置 CentOS 虚拟机-网络适配器为“桥接模式(直接连接物理网络),复制物理网络连接状态” [步骤02]配置虚拟网卡 [步骤03]配置 CentOS 网络 [步骤04]测试 windo ...