生成元（Digit Generator,ACM/ICPC Seoul 2005, UVa1583）

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

216

121

2005

Sample Output

198

1979

分析：

假设所求生成元为m.m<n。只需枚举所有的m<n，看看有没有哪个数是n的生成元
  一次性枚举内的所有正整数m，标记“m加上m的各个数字之和得到的数有一个生成元是m”
  第二种效率比第一种高，因为第一种每次计算一个 n的生成元都需要枚举n-个数。
           

#include <stdio.h>
#include <string.h>

#define maxn 100005
int a[maxn];

int main()
{
    int T, n;
    memset(a, , sizeof(a));
    for(int m = ; m < maxn; m++){
        int x = m, y = m;
        while(x > ){
            y += x % ;
            x /= ;
        }
        if(a[y] ==  || m < a[y])
            a[y] = m;
    }
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        printf("%d\n", a[n]);
    }
    return ;
}