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.

Sample Input

4 5
1 2 3 7 5
3 6
7 11 2 5 13 17
0 0

Sample Output

3 5
2 3 4

和另一道题有相似的地方

但是修改过代码再提交一直WA

看了别人的代码,样例输出和我写的是一样的但是不知道为什么,我写的一直WA

 1 #include<iostream>
2 #include<cstdlib>
3 #include<cstdio>
4 #include<cstring>
5 #include<algorithm>
6 #include<cmath>
7 using namespace std;
8 int a[100000] , mod[100000] ;
9 int main()
10 {
11 int c , n ;
12 while ( scanf("%d%d",&c,&n) , c || n )
13 {
14 int i , j ;
15 for ( i = 0 ; i < n ; i ++ )
16 scanf("%d",&a[i]) , mod[i] = -2 ;//将mod初始化为-2
17 mod[0]=-1 ;//mod[0]为-1,就是假设存在a[-1],且a[-1]是n的倍数,这样就可以把两种情况写在一起
18 __int64 sum = 0 ;//直接用sum,省去了另开数组的空间
19 for ( i = 0 ; i < n ; i ++ )
20 {
21 sum += a[i] ;
22 if ( mod [ sum % c ] != -2 )
23 {//如果在i之前有与sum对n同余的数,则可以输出答案,
24 for ( j = mod [ sum % c ] + 1 ; j <= i ; j ++ )
25 {
26 cout<<j+1;
27 if ( i != j )
28 cout<<' ';
29 }
30 cout<<endl;
31 break;
32 }
33 mod [sum%c] = i ;//记录余数对应的是i
34 }
35 }
36 return 0;
37 }

C - 抽屉 POJ - 3370 (容斥原理)的更多相关文章

  1. B - 抽屉 POJ - 2356 (容斥原理)

    The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers ...

  2. 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 ...

  3. POJ 3370 Halloween treats(抽屉原理)

    Halloween treats Every year there is the same problem at Halloween: Each neighbour is only willing t ...

  4. POJ 3370 Halloween treats(抽屉原理)

    Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6631   Accepted: 2448 ...

  5. poj 2773(容斥原理)

    容斥原理入门题吧. Happy 2006 Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 9798   Accepted: 3 ...

  6. Poj 3370

    题目传送门:https://vjudge.net/problem/POJ-3370 题意:在n个数中找K个数使得他们的和为c的倍数. 题解:抽屉原理,同poj 2356 只不过写法上有所简化. 简化版 ...

  7. [POJ 3370] Halloween treats

    Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7143   Accepted: 2641 ...

  8. poj 3370 鸽笼原理知识小结

    中学就听说过抽屉原理,可惜一直没机会见识,现在这题有鸽笼原理的结论,但其实知不知道鸽笼原理都可以做 先总结一下鸽笼原理: 有n+1件或n+1件以上的物品要放到n个抽屉中,那么至少有一个抽屉里有两个或两 ...

  9. POJ 2356 && POJ 3370 鸽巢原理

    POJ 2356: 题目大意: 给定n个数,希望在这n个数中找到一些数的和是n的倍数,输出任意一种数的序列,找不到则输出0 这里首先要确定这道题的解是必然存在的 利用一个 sum[i]保存前 i 个数 ...

随机推荐

  1. Centos7 升级 sqlite3

    下载地址:https://www.sqlite.org/download.html [root@djangoServer ~]# wget https://www.sqlite.org/2019/sq ...

  2. 注册 Amazon Web Services(AWS) 账号,助园一臂之力

    感谢大家去年的大力支持,今年园子继续和 Amazon Web Services(AWS) 合作,只要您通过 博客园专属链接 注册一个账号(建议使用手机4G网络注册),亚马逊就会给园子收入,期待您的支持 ...

  3. .NET Core Generic Host项目使用Topshelf部署为Windows服务

    1..NET Core Generic Host是什么? 在.NET Core 2.1版本加入了一种新的Host,即Generic Host(通用主机). 现在在2.1版本的Asp.Net Core中 ...

  4. 漏洞复现-CVE-2018-15473-ssh用户枚举漏洞

          0x00 实验环境 攻击机:Win 10 0x01 影响版本 OpenSSH 7.7前存在一个用户名枚举漏洞,通过该漏洞,攻击者可以判断某个用户名是否存在于目标主机 0x02 漏洞复现 针 ...

  5. 《Asp.Net Core3 + Vue3入坑教程》 - 6.异常处理与UserFriendlyException

    简介 <Asp.Net Core3 + Vue3入坑教程> 此教程适合新手入门或者前后端分离尝试者.可以根据图文一步一步进操作编码也可以选择直接查看源码.每一篇文章都有对应的源码 目录 & ...

  6. 越来越受欢迎的Vue想学么,90后小姐姐今儿来教你

    摘要:Vue的相关技术原理成为了前端岗位面试中的必考知识点,掌握 Vue 对于前端工程师来说更像是一门"必修课". 本文原作者为尹婷,擅长前端组件库研发和微信机器人. 我们发现, ...

  7. MindSpore:基于本地差分隐私的 Bandit 算法

    摘要:本文将先简单介绍Bandit 问题和本地差分隐私的相关背景,然后介绍基于本地差分隐私的 Bandit 算法,最后通过一个简单的电影推荐场景来验证 LDP LinUCB 算法. Bandit问题是 ...

  8. VUE中的子父组件、兄弟组件之间相互传值,相互调用彼此的方法

    vue--组件传值 父组件传值给子组件--"props" 一.父组件--示例 <template> <child :choose-data="choos ...

  9. 如何快速的插入 100W数据到数据库,使用PreparedStatement 最快实现!

    有时候,我们使用数据库的时候,如何快速的添加测试数据到数据库中,做测试呢,添加100W 数据,如果使用工具的话可能很慢,这里我推荐大家使用 PreparedStatement 预编译 去进行操作:单线 ...

  10. C# 基础 - 日志捕获二使用 log4net

    引入 log4net.dll 项目->添加->新建项->应用程序配置文件,命名为 log4net.config,并把属性的复制到输出目录设置为 如果较新则复制,后续客户端需要读取在 ...