主题链接：
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 a generating 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 < 10 5).
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 10 5. If there are several generating sets, output any of them. If there are no generating sets, output −1.
Sample
| input | output |
|---|---|
17 | 37 12 43 |
代码例如以下：
#include#include #include #include using namespace std;const int maxn = 100017;vector 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;}
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