SGU 108 Self-numbers 2

108. Self-numbers 2

time limit per test: 1 sec. memory limit per test: 4096 KB

 

In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence 33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ... The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with no generators is a self-number. Let the a[i] will be i-th self-number. There are thirteen self-numbers a[1]..a[13] less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97. (the first self-number is a[1]=1, the second is a[2] = 3, :, the thirteen is a[13]=97);

 

Input

Input contains integer numbers N, K, s₁...s_k. (1<=N<=10⁷, 1<=K<=5000) delimited by spaces and line breaks.

 

Output

At first line you must output one number - the quantity of self-numbers in interval [1..N]. Second line must contain K numbers - a[s₁]..a[s_k], delimited by spaces. It`s a gaurantee, that all self-numbers a[s₁]..a[s_k] are in interval [1..N]. (for example if N = 100, s_k can be 1..13 and cannot be 14, because 14-th self-number a[14] = 108, 108 > 100)

 

Sample Input

100 10
1 2 3 4 5 6 7 11 12 13

Sample Output

13
1 3 5 7 9 20 31 75 86 97

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 5005
#define ll long long
#define INF 0x7fffffff
#define eps 1e-8
pair<int, int>a[maxn];
int n,m,k,t,x,pos;
bool v1[64], v2[64];
int ans[maxn];
int main(){
    while (~scanf("%d%d",&n,&k)){
        for (int i = 0; i < k; i++){ scanf("%d", &a[i].first); a[i].second = i; }
        sort(a, a + k); 
        pos = 0;t = 0;
        memset(v1, true, sizeof v1);
        memset(v2, true, sizeof v2);
        for (int i = 1; i <= n; i++){
            if (i % 64 == 0){
                memcpy(v1, v2,64);
                memset(v2, true, sizeof v2);
            }
            if (v1[i % 64]){
                t++;
                while (a[pos].first == t)ans[a[pos++].second] = i;
            }
            m = 0;x = i;
            while (x){
                m += x % 10;
                x /= 10;
            }
            if (m + i % 64 >= 64)v2[(m + i % 64)%64] = false;
            else v1[m + i % 64] = false;
        }
        printf("%d\n", t);
        for (int i = 0; i < k; i++)printf(i == k - 1 ? "%d\n" : "%d ", ans[i]);
    }
    return 0;
}