B - 抽屉 POJ - 2356 (容斥原理)
原理 poj 容斥 抽屉
2023-09-11 14:22:48 时间
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.
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
题意:
输入n
输入n个数
判断这n个数中是不是有几个数字之和是n的倍数
思路:
n个数余数分别为 1 ~ n-1 ,相当于有n-1个抽屉,n个物品
分别计算a[1] + a[2] + …… + a[k] 的和然后取余如果为零则直接输出前k个数,否则寻找余数相同的两个数,假设为i, j (i < j),则a[i+1] + . . . . + a[j] 的和一定能被n整除(原理还没想清楚)
AC代码
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 using namespace std; 5 int a[10005]; 6 int mod[10005]; 7 int mark[10005]; 8 9 int main() 10 { 11 int n; 12 bool flag = false; 13 cin >> n; 14 memset(mod, 0, sizeof(mod)); 15 memset(mark, 0, sizeof(mark)); 16 for(int i = 1; i <= n; i++) 17 { 18 cin >> a[i]; 19 mod[i] = (mod[i-1] + a[i]) % n; 20 } 21 22 for(int i = 1; i <= n; i++) 23 { 24 if(mod[i] == 0) 25 { 26 flag = true; 27 cout << i << endl; 28 for(int j = 1; j <= i; j++) 29 cout << a[j] << endl; 30 break; 31 } 32 } 33 34 if(!flag) 35 { 36 for(int i = 1; i <= n; i++) 37 { 38 if(mark[mod[i]] == 0) 39 mark[mod[i]] = i; 40 else 41 { 42 cout << i -mark[mod[i]] << endl; 43 for(int j = mark[mod[i]]+1; j <= i; j++) 44 cout << a[j] << endl; 45 46 break; 47 } 48 } 49 } 50 51 return 0; 52 }
相关文章
- POJ 3370 Halloween treats 鸽巢原理 解题
- 初涉IPC,了解AIDL的工作原理及用法
- redux 中间件的原理是什么?
- 信号源结构原理
- jQuery技术内幕:深入解析jQuery架构设计与实现原理. 3.1 总体结构
- DNS攻击原理与防范
- 《精通Python网络爬虫:核心技术、框架与项目实战》——第二篇 Part 2核心技术篇 第3章 网络爬虫实现原理与实现技术 3.1 网络爬虫实现原理详解
- RocketMQ详解(三)启动运行原理
- 浅析HTTPS原理:网络通信的3大问题、对称加密算法、非对称加密算法、对称+非对称交换密钥具体流程及安全隐患、数字证书如何生成颁发如何验证、证书链、hash算法比较摘要防止篡改的流程、HMAC消息认证码对hash的优化、HTTPS的宏观安全模型、安全连接建立流程
- 浏览器工作原理:浅析从输入URL到页面展示这中间发生了什么
- 秋色园QBlog技术原理解析:Module之页面基类设计(五)
- Hessian 原理分析--转
- 深入了解SQLServer系统数据库工作原理
- Socket通信原理