Digit Generator UVA - 1583

​ 　 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’.

Sample Input

3
216
121
2005

Sample Output

198
0
1979

HINT

​ 　 题目的意思很明显，要求满足256 (= 245 + 2 + 4 + 5)形式的整数。一个n位整数假设每一位都是9，那么要求的数里面可能的最小的数就是N-9*n;（其中N是给定的数字）。

Accepted

#include<stdio.h>

int min(int t)			//求出给定数字的位数
{					   //也可以用循环来求出
    if(t<10)return 1;
    if(t<100)return 2;
    if(t<1000)return 3;
    if(t<10000)return 4;
    if(t<100000)return 5;
    if(t<1000000)return 6;
}

int mm(int t)		//求出每一位数字的和
{
    int k=0;
    while(t)
    {
        k+=t%10;
        t/=10;
    }
    return k;
}

int main()
{
    int sum;
    scanf("%d",&sum);
    while(sum--)
    {
        int flag=0;
        int t;
        scanf("%d",&t);
        int k=min(t);
        int minst=t-k*9;
        minst=minst>0?minst:1;
        for(minst;minst<t;minst++)
        {
            if((minst+mm(minst))==t)
            {
                printf("%d\n",minst);
                continue;
            }
        }
        printf("0\n");

    }
}