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
- 【机器学习学习】SKlearn + XGBoost 预测 Titanic 乘客幸存
Titanic 数据集是从 kaggle下载的,下载地址:https://www.kaggle.com/c/titanic/data 数据一共又3个文件,分别是:train.csv,test.csv, ...
- Java异常(输出[D@139a55问题)
简单给出一段代码说明问题: public class Main { public static void main(String[] args) { double a[]={1,4,3,2}; Sys ...
- JavaScript splice() 方法
定义和用法 splice() 方法向/从数组中添加/删除项目,然后返回被删除的项目. 注释:该方法会改变原始数组. 例子 1 在本例中,我们将创建一个新数组,并向其添加一个元素: <script ...
- hdu_1370Biorhythms(互素的中国剩余定理)
Biorhythms Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...
- HDU-1013九余数定理
题目传送门:HDU1013 九余数定理 //题目描述:给定一个数,要求你求出它的每位上的数字之和,并且直到每位上的数字之和为个位时候输出它 //输入:一个整数 //输出:题目描述的结果 //算法分析: ...
- 解决在SecurecCRT登录后,发现方向键、backspace(退格键)、delete(删除键)为乱码的问题
问题:使用securecrt ssh到linux之后,backspace(退格键),delete(删除键),以及4个方向键都为乱码,不能正常使用.按tab键也没有自动补全文件名. 即: 按Backsp ...
- html日历(1)
<html> <head> <link rel="stylesheet" type="text/css" href="S ...
- Spark算子--union、intersection、subtract
转载请标明出处http://www.cnblogs.com/haozhengfei/p/252bcc1d1ab30c430d347279d5827615.html union.intersection ...
- EntityFramework默认映射规则
我不太习惯通过CodeFirst去维护数据库(尽管这是未来实现自动编程的必经之路),还是喜欢通过数据库设计工具如PowerDesigner去建表.如果不想对EF的实体和数据表做什么映射的话,就要注意默 ...
- 【开发技术】Eclipse设置软tab(用4个空格字符代替)及默认utf-8文件编码(unix)
Eclipse设置软tab(用4个空格字符代替)及默认utf-8文件编码(unix) 本文摘要: 1.如何配置Eclipse中编辑器支持softtab(用数个空格字符代替默认的tab缩进): 2.如何 ...