Uva - 1583 - Digit Generator

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 Tis given in the first line of the input. Each test case takes one line containing an integer N , 1N100, 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

AC代码：

#include <stdio.h>#include <string.h>const int MAXN = 100005;int ans[MAXN]; // 数组比较大的情况下，就将数组放在堆中int main(){int T, n;memset(ans, 0, sizeof(ans));// 先用循环遍历，枚举出100000以内的所有正整数的最小生成元for (int m = 1; m < MAXN; m++) {int x = m, y = m;while (x) {y += x % 10;x /= 10;}if ((ans[y] == 0) || m < ans[y]) {ans[y] = m;}}scanf("%d", &T);// 最后这里直接查表即可while (T--) {scanf("%d", &n);printf("%d\n", ans[n]);}return 0;}