URAL 1727. Znaika's Magic Numbers（数学 vector）

主题链接：http://acm.timus.ru/problem.aspx?space=1&num=1727

1727. Znaika's Magic Numbers

Time limit: 0.5 second

Memory limit: 64 MB

Znaika has many interests. For example, now he is investigating the properties of number sets. Znaika writes down some set consisting of different positive integers (he calls this set agenerating set),
calculates the sum of all the written digits, and writes down the result in a special notebook. For example, for a generating set 7, 12, 43, he will write down the number17 = 7 + 1 + 2 + 4 + 3. Znaika is sure that only magic numbers can
appear as a result of this operation.

Neznaika laughs at Znaika. He thinks that there is a generating set for every number, and he even made a bet with Znaika that he would be able to construct such a set.

Help Neznaika win the bet and construct a generating set for a given number.

Input

The only input line contains an integer n (0 < n < 105).

Output

If it is possible to construct a generating set for the number n, output the number of elements in this set in the first line. In the second line output a space-separated list of these elements.
The elements of the set must be different positive integers strictly less than 105. If there are several generating sets, output any of them. If there are no generating sets, output −1.

Sample

input	output
17	3 7 12 43

代码例如以下：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 100017;
vector<int> vv[maxn];
int vis[maxn], ans[maxn];
void init()
{
    for(int i = 1; i < maxn; i++)
    {
        int tt = i;
        int sum = 0;
        while(tt)
        {
            sum+=tt%10;
            tt/=10;
        }
        vv[sum].push_back(i);
    }
}
int main()
{
    int n;
    init();
    while(~scanf("%d",&n))
    {
        memset(vis, 0, sizeof(vis));
        int k = 0;
        for(int i = 45; i > 0; i--)
        {
            if(n >= i)
            {
                for(int j = 0; j < vv[i].size(); j++)
                {
                    if(n < i)
                        break;
                    if(vis[vv[i][j]] == 0)
                    {
                        vis[vv[i][j]] = 1;
                        ans[k++] = vv[i][j];
                        n-=i;
                    }
                }
            }
            if(n <= 0)
                break;
        }
        if(k==0 || n != 0)
        {
            printf("-1\n");
            continue;
        }
        printf("%d\n%d",k,ans[0]);
        for(int i = 1; i < k; i++)
        {
            printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。