Codeforces Round #441 (Div.2) - C - Classroom Watch

http://codeforces.com/problemset/problem/875/A

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 x written 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.

Input

The first line contains integer n (1 ≤ n ≤ 109).

Output

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.

Examples

input

21

output

1
15

input

20

output

0

Note

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.

题意

直接看题目也是没有看出题意来，幸亏有Note
应该看Note都能看出题意来吧

题解

一种是转换成字符串，一种是直接计算，由于题意和方法都很简单就不过多赘述
只是一些小技巧还是可以记一下的，比如怎样判断整数位数、字符和整数来回转换

字符串计算

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int ans[1000000];
int main()
{
    char ch[11];
    int n;
    scanf("%d",&n);
    itoa(n, ch, 10);
    int ll=strlen(ch);
    int num=0;
    for(int gg=n-ll*9;gg<=n;gg++)
    {
        itoa(gg, ch, 10);
        int l=strlen(ch);
        long long sum=gg;
        for(int i=0;i<l;i++)
            sum+=(ch[i]-'0');
        if(sum==n)
        {
            ans[num]=gg;
            num++;
        }
    }
        printf("%d\n",num);
        for(int i=0;i<num;i++)
            printf("%d\n",ans[i]);
}

整数计算

#include <bits/stdc++.h>  
using namespace std;  
int n,a[10010];  
int main(){  
    while(~scanf("%d",&n))
    {  
        int temp = n, cnt = 0;  
        while(temp)
        {  
            cnt ++; 
            temp /= 10;  
        }  
        int ans = 0;  
        for(int i = n-cnt * 9; i <= n; i ++)
        {  
            int j = i, sum = i;  
            while(j)
            {  
                sum += (j % 10);  
                j /= 10;  
            }  
            if(sum == n) a[ans ++] = i;  
        }  
        printf("%d\n",ans);  
        for(int i = 0; i < ans; i ++)
        {  
            if(!i) printf("%d",a[i]);  
            else printf(" %d",a[i]);  
        }  
        if(ans) printf("\n");  
    }  
    return 0;  
}