poj2356
来源:互联网 发布:淘宝时间和北京时间差 编辑:程序博客网 时间:2024/05/18 14:23
Find a multiple
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.
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
512341
Sample Output
223
Source
Ural Collegiate Programming Contest 1999
鸽巢原理。对于给定的序列a1,a2,a3,...an,存在一个连续子序列ai~aj,使得其和整除n。与poj3370几乎完全相同。
代码如下:
#include<stdio.h>#include<string.h>int main(void) { int number[10001]; int left[10001]; int rem; int n; int start; int end; while (scanf("%d", &n) != -1) { memset(left, 0, sizeof (left)); rem = 0; for (int i = 1; i <= n; i++) scanf("%d", &number[i]); for (int i = 1; i <= n; i++) { rem = (rem + number[i]) % n; if (rem == 0) { start = 1; end = i; break; } if (left[rem]) { start = left[rem] + 1; end = i; break; } left[rem] = i; } printf("%d\n", end - start + 1); for (int i = start; i <= end; i++) printf("%d\n", number[i]); } return 0;}
- poj2356
- poj2356
- poj2356
- poj2356
- poj2356 二分
- poj2356(鸽巢原理)
- poj2356 抽屉原理
- poj2356(二分图匹配)
- poj2356——Find a multiple
- poj3370 poj2356 鸽巢定理
- poj2356 Find a multiple(鸽巢原理)
- poj2356 Find a multiple(鸽巢原理)
- POJ2356 Find a multiple【鸽巢原理】
- poj2356 Find a multiple 抽屉原理
- 【POJ2356】Find a multiple(鸽巢原理)
- 【HDU1205/POJ2356/POJ3370】鸽巢原理专题
- 鸽巢原理 Poj3370&Hdu1808 + Poj2356 + Hdu 1205
- POJ2356 Find a multiple 抽屉原理(鸽巢原理)
- Spring对Hibernate的事务支持
- UVa 10004 - Bicoloring
- window 计时器IO监控
- (一、预备知识)2、ios系统发展历史和前景
- 八个移动产品设计必备网站
- poj2356
- int、short、char 类型超出范围赋值__说说原码,反码,补码
- Win7+Ubuntu11.10(EasyBCD硬盘安装)
- linux sed命令参数及用法详解---linux 利用script来处理文本文件
- 应届生工作后的迷茫...
- 关于VC6.0调试和重装的一点心得
- logmner
- IO基本概念
- Working with Alternative Menus