鸽巢原理应用-分糖果 POJ 3370 Halloween treats
基本原理:n+1只鸽子飞回n个鸽笼至少有一个鸽笼含有不少于2只的鸽子。
很简单,应用却也很多,很巧妙,看例题:
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 containsn 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 aisweets). 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
Source
题目大意:糖果平分问题。有c个小孩,n个提供糖果的邻居,你可以选择要或不要。现在你只考虑得到的全部糖果能否平分,可能有多种方案,输出一种即可。
上面的case 1: 结果 2 3 4 也行,总和为12. 输出一种即可
#include <stdio.h>
#include <algorithm>
using namespace std; int c,n,neigb[100001];
int S;
struct Remnant
{
int h,r; // 下标和余数
}R[100001]; bool cmp(const Remnant &a, const Remnant & b){ //按余数从小到大排序
if( a.r == b.r)
return a.h < b.h;
return a.r < b.r;
} int main(){
//freopen("in.txt","r",stdin);
while(scanf("%d %d", &c, &n) != EOF){
if(c==0 && n==0) break;
int k=-1,h;
S=0;
for(int i=0; i<n; i++)
{
scanf("%d",&neigb[i]);
S += neigb[i];
R[i].r = S%c; //存储是前i个和 对c的余数
R[i].h = i + 1; //h 为下标
if(k == -1 && R[i].r==0 ) k=i;
} if(k == -1){
sort(R, R+n, cmp);
for(int i=0; i<n-1; i++)
{
if(k == -1 && R[i].r == R[i+1].r)
{
k = R[i].h;
h = R[i+1].h;
break;
}
}
if(k==-1)
printf("no sweets\n");
else{
for(int i=k+1; i<h; i++)
printf("%d ",i);
printf("%d\n",h);
}
}else{
for(int i=0; i<k; i++)
printf("%d ",i+1);
printf("%d\n",k+1);
} }
return 0;
}
鸽巢原理应用-分糖果 POJ 3370 Halloween treats的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- [POJ 3370] Halloween treats
Halloween treats Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7143 Accepted: 2641 ...
- 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 ...
- POJ 3370 Halloween treats(抽屉原理)
Halloween treats Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6631 Accepted: 2448 ...
- 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 ...
- 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 ...
- 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 ...
随机推荐
- 【deep learning学习笔记】注释yusugomori的LR代码 --- LogisticRegression.h
继续看yusugomori的代码,看逻辑回归.在DBN(Deep Blief Network)中,下面几层是RBM,最上层就是LR了.关于回归.二类回归.以及逻辑回归,资料就是前面转的几篇.套路就是设 ...
- 邮件应用Acompli和日历应用Sunrise(传微软曾考虑以80亿美元收购企业通讯公司Slack)
http://tech.163.com/16/0305/10/BHCU8EHO000915BD.html http://www.cnbeta.com/articles/480835.htm
- CImageList使用简要说明
CImageList ImageList;//创建一个包含3个24位色32x32图片的ImageList,ILC_MASK的意思是同时创建一个mask,这样在下面指定了背景颜色以后ImageList就 ...
- 14.2.5.4 Physical Structure of an InnoDB Index InnoDB Index 的物理结构
14.2.5.4 Physical Structure of an InnoDB Index InnoDB Index 的物理结构 所有的InnoDB indexes 是B-trees ,index ...
- Shell 传递参数
Shell 传递参数 向脚本传递参数,格式为:$n. 向脚本传递三个参数,并分别输出: echo "Shell 传递参数实例!"; echo "第一个参数为:$1&quo ...
- mysql 主从同步出问题,重新修复从库 - web架构研究
mysql 主从同步出问题,重新修复从库 - web架构研究 mysql 主从同步出问题,重新修复从库 0 昨天由于操作失误,在从库上执行一堆sql之后,导致主从同步错误,并且已 ...
- 改动已有gpg密钥的用户标识及凝视
/********************************************************************* * Author : Samson * Date ...
- uva 1415 - Gauss Prime(高斯素数)
题目链接:uva 1415 - Gauss Prime 题目大意:给出一个a,b,表示高斯数a+bi(i=−2‾‾‾√,推断该数是否为高斯素数. 解题思路: a = 0 时.肯定不是高斯素数 a != ...
- 显示形状回归算法(ESR)代码介绍
源地址:http://www.thinkface.cn/thread-3704-1-6.html 人脸对齐包括两个部分,分别为训练部分和测试部分.所有的代码基于opencv2.0.(一)训练阶段Ste ...
- CentOS6-釋放ip重新分配,centos7 ifconifg没有ip
http://bbs.csdn.net/topics/390725823 系统win7 ,dhcp自动获取ip虚拟机是10.0 安装之后我装了ubuntu 选用 NAT网络, 刚装完我能上网 ,但是 ...