Codeforces C. Classroom Watch
1 second
512 megabytes
standard input
standard output
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number n. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that n is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer x was given. The task was to add x to the sum of the digits of the number xwritten in decimal numeral system.
Since the number n on the board was small, Vova quickly guessed which x could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number n for all suitable values of x or determine that such x does not exist. Write such a program for Vova.
The first line contains integer n (1 ≤ n ≤ 109).
In the first line print one integer k — number of different values of x satisfying the condition.
In next k lines print these values in ascending order.
21
1
15
20
0
In the first test case x = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such x.
从n-n的位数*9到n枚举就行了;
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std; int n,ans[],res,k; int main(){
scanf("%d",&n);
long long g=,gg=;
while(g<=n){
g=g*+;
gg++;
}
gg+=;
for(int i=n-gg*;i<=n;i++){
int x=n-i,a=i;
while(a>){
x=x-a%;
a=a/;
}
if(x==){
res++;
ans[res]=i;
}
}
printf("%d\n",res);
for(int i=;i<=res;i++)
printf("%d\n",ans[i]);
}
Codeforces C. Classroom Watch的更多相关文章
- Codeforces 876C Classroom Watch:枚举
题目链接:http://codeforces.com/contest/876/problem/C 题意: 定义函数:f(x) = x + 十进制下x各位上的数字之和 给你f(x)的值(f(x) < ...
- CodeForces - 876C Classroom Watch (枚举)
题意:已知n,问满足条件"x的各个数字之和+x=n"的x有几个并按升序输出. 分析: 1.n最大1e9,10位数,假设每一位都为9的话,可知x的各个数字之和最大可以贡献90. 2. ...
- codeforces 876 C. Classroom Watch
http://codeforces.com/contest/876/problem/C C. Classroom Watch time limit per test 1 second memory l ...
- Codeforces Round #561 (Div. 2) A. Silent Classroom
链接:https://codeforces.com/contest/1166/problem/A 题意: There are nn students in the first grade of Nlo ...
- Codeforces Round #441 (Div. 2, by Moscow Team Olympiad) C. Classroom Watch
http://codeforces.com/contest/876/problem/C 题意: 现在有一个数n,它是由一个数x加上x每一位的数字得到的,现在给出n,要求找出符合条件的每一个x. 思路: ...
- Codeforces 1166A - Silent Classroom
题目链接:http://codeforces.com/problemset/problem/1166/A 思路:统计所有首字母出现的次数,由贪心可知对半分最少. AC代码: #include<i ...
- Codeforces Round #561 (Div. 2) A. Silent Classroom(贪心)
A. Silent Classroom time limit per test1 second memory limit per test256 megabytes inputstandard inp ...
- codeforces Round #441 C Classroom Watch【枚举/注意起点】
C. time limit per test 1 second memory limit per test 512 megabytes input standard input output stan ...
- 「Codeforces Round #441」 Classroom Watch
Discription Eighth-grader Vova is on duty today in the class. After classes, he went into the office ...
随机推荐
- Grafana最新版本4.3.1安装(后端使用mysql)
环境 CentOS release 6.5 (Final) 64bitzabbix_server (Zabbix) 3.0.3 grafana-4.3.1mysql-5.6.21 一.安装grafan ...
- Angular-搜索框及价格上下限
Angular-搜索框及价格上下限 闲来无事,写一个简单的angular的搜索框. 1.要求: 利用 AngularJS 框架实现手机产品搜索功能,题目要求: 1)自行查找素材,按照原有数据格式将手机 ...
- Elixir的Phoenix框架:请求处理之道
本文基于Phoenix1.3,但请求的处理流程跟1.2基本一致,只是模块的命名和目录结构有所差异. 简单介绍,phoenix是一个网站框架,本质就是http请求处理.这篇文章主要就是讲一个请求,在结果 ...
- struts2中配置文件的调用顺序
1.default.properties 该文件保存在 struts2-core-2.3.7.jar 中 org.apache.struts2中 2.struts-default.xml 该文件保存在 ...
- 读书笔记《PHP与MySQL程序设计》一
第1章 PHP概述 1.1 历史(PHP4.PHP5.PHP5.3.PHP6[未发布]) 1.2 一般语言特性(实用性.强大功能.可选择性.成本[开源]) 第2章 环境配置 2.1 安装的前提条件( ...
- hdu_2087 剪花布条(kmp)
剪花布条 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...
- An Easy Problem?!(细节题,要把所有情况考虑到)
http://poj.org/problem?id=2826 An Easy Problem?! Time Limit: 1000MS Memory Limit: 65536K Total Sub ...
- zookeeper的安装以及启动jps进程
2.7.1安装 将下载好的安装包,解压到指定位置,这里为直接解压到当前位置,命令如下: tar -zxvf zk-{version}.tar.gz 修改zk配置,将zk安装目录下conf/zoo_sa ...
- 程序员听到bug后的N种反应,太形象了
程序员的世界里,不止有代码,还有bug,bug,bug 当出现bug时,程序员们的反应是怎样的呢?
- 制作ssh互信的docker镜像
Dockerfile FROM ubuntu:16.04 # package RUN apt-get update; apt-get -y install ssh COPY ssh_config /e ...