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. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line.

Sample Output

1
3
5
7
9
20
31
42
53
64
|
| <-- a lot more numbers
|
9903
9914
9925
9927
9938
9949
9960
9971
9982
9993
|
|

找出1000000以內所有不能由其他數字（abc+a+b+c）組成的數

因為a+b+c+...最大也就是9+9+9+9+9+9=54，所以直接暴力即可

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define ll long long
using namespace std;
int main(){
	for(int i=1;i<=1000000;++i){
		int u=0;
		for(int j=1;j<=54&&i-j>0;++j){
			int p=i-j,s=0;
			while(p>0){
				s+=p%10;
				p/=10;
			}
			if(s==j){
				u=1;
				break;
			}
		}
	  	if(u==0)
			cout<<i<<endl;
	}
	return 0;
}

#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
int hash[1000005];
void deal(int n){
    int rec = n;
    while( n > 0 ){
        int c = n % 10;
        n /= 10;
        rec += c;
    }
    hash[rec] = 1;
}
int main(){
    for(int i=1;i<=1000000;++i){
        if(!hash[i])
            printf("%d\n",i);
        deal(i);
    }
    return 0;
}