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
Output for the Sample Input
198 0 1979
要求是根据输入的数找到其最小生成元，我一开始的思路是枚举，但量太多，想用判断来缩小枚举范围。设输入数num，当num=99999时的各个位数的最大和为45，也就是说只要在[num-45,num]这个区间进行枚举，就可以比较轻松地得到答案。以下用C写的答案：
#include<stdio.h>
int main()
{int ans=0,turns,num,i;
scanf("%d",&turns);while(turns>0){
scanf("%d",&num);
for(i=num-45;i<num;i++)
{if((i+i%10000/1000+i%1000/100+i%100/10+i%10)==num)
{ans=i; break;}}
if(!ans)printf("No answers\n");
else printf("%d\n",ans);turns--;ans=0;
}
return 0;
}