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
- Nginx配置文件(2)
一.配置文件结构 1.全局块:配置影响nginx全局的指令.一般有运行nginx服务器的用户组,nginx进程pid存放路径,日志存放路径,配置文件引入,允许生成worker process数等. 2 ...
- 学习笔记-express路径问题
在页面渲染成功之后,报错出现静态文件css样式引用路径出错,于是我就根据express api文档,托管静态文件作出修改,最后全是徒劳.于是我又从引用开始找起,<link rel="s ...
- hackerrank Alex对战Fedor
任意门 为了在漫长得飞行旅途中娱乐,Alex和Fedor发明了如下的一个简单的双人游戏.游戏是: 首先, Alex画一个有权无向图.该图中可能有多重边(多重边的权值可能相同或者不同). 然后,Fedo ...
- bzoj:1703: [Usaco2007 Mar]Ranking the Cows 奶牛排名
Description 农夫约翰有N(1≤N≤1000)头奶牛,每一头奶牛都有一个确定的独一无二的正整数产奶率.约翰想要让这些奶牛按产奶率从高到低排序. 约翰已经比较了M(1≤M≤100 ...
- vijos 1213:80人环游世界
描述 想必大家都看过成龙大哥的<80天环游世界>,里面的紧张刺激的打斗场面一定给你留下了深刻的印象.现在就有这么一个80人的团伙,也想来一次环游世界. 他们打算兵分多路,游遍每一个国家. ...
- HDU_4883
TIANKENG's restaurant Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/O ...
- 常用Windows DOS命令项目部署经常用到
img { max-width: 100% } 前两天部署.netcore项目,首先是生产环境域名访问不了,再到.netcore项目IIS部署502.5,在到莫名其妙的500,在排查项目部署问题的时候 ...
- [国嵌笔记][029][ARM处理器启动流程分析v2]
2440启动流程 启动方式:nor flash启动.nand flash启动 地址布局: 选择nor flash启动时,SROM(nor flash)地址为0x00000000 选择nand flas ...
- java中JFrame类中函数addWindowListener(new WindowAdapter)
转自:http://blog.csdn.net/datouniao1/article/details/46984987:侵删. 在java编写的过程中常常遇到样的一段代码: frame.addWind ...
- 利用nginx 虚拟主机、请求转发实现不同端口web访问
一个服务器上挂一个网站实在是有点浪费:一个服务器上可以放多个网站:可以开启nginx的虚拟主机功能:利用访问的路径或者域名不同访问不同的文件夹:例如: 1.一台服务器上放多个网站使用nginx的配置文 ...