UVa 1583 - Digit Generator

https://uva.onlinejudge.org/external/15/1583.pdf

For a positive integer N , the digit-sum of N is defined as the sum of N itself and its digits. When M is the
digitsum of N , we call N a generator of M .
For example, the digit-sum of 245 is 256 (= 245 + 2 + 4 + 5). Therefore, 245 is a generator of
256.
Not surprisingly, some numbers do not have any generators and some numbers have more than one generator.
For example, the generators of 216 are 198 and 207.
You are to write a program to find the smallest generator of the given integer.
Input
Your program is to read from standard input. The input consists of T test cases. The number of test cases T is
given in the first line of the input. Each test case takes one line containing an integer N , 1 N 100, 000 .
Output
Your program is to write to standard output. Print exactly one line for each test case. The line is to contain a
generator of N for each test case. If N has multiple generators, print the smallest. If N does not have any
generators, print 0.
The following shows sample input and output for three test cases.
Sample Input
3
216
121
2005
Sample Output
198
0
1979
Seoul 2005-2006

题意：
如果x加上x的各位数字之和得到y，那么就说x是y的生成元，现在给你一个N，让你求N的最小生成元。

#include <iostream>#include <sstream>#include <iomanip>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <bitset>#include <string>#include <numeric>#include <algorithm>#include <functional>#include <iterator>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <complex>#include <ctime>#define INF 0x3f3f3f3f#define eps 1e-6#define p(x) printf("%d\n", x)#define k(x) printf("Case %d: ", ++x)#define mes(x, d) memset(x, d, sizeof(x))#define s(x) scanf("%d", &x)typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;int ans[100005];int main(){    //freopen("int.txt","r",stdin);    //freopen("out.txt","w",stdout);    memset(ans,0,sizeof(ans));    for(int i = 1;i < 100005;i++)    {        int x = i,y = i;        while(x > 0)        {            y += x % 10;            x /= 10;        }        if(ans[y] == 0 || i < ans[y])            ans[y] = i;    }    int T,N;    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        printf("%d\n",ans[N]);    }    return 0;}