uva 1583 Digit Generator

原題：
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

中文：

給你一個數x，如果x可以由y加上y的每一位得到，那麼y是x的產生元~
現在給你一個數n從1到100000，問你這個數最小的產生元是多少。

#include <bits/stdc++.h>using namespace std;typedef long long ll;int solve(int x){    int ans=0;    while(x)    {        ans+=x%10;        x/=10;    }    return ans;}int main(){    ios::sync_with_stdio(false);    int t,n;    cin>>t;    while(t--)    {        cin>>n;        int tmp=n,dig=0;        while(tmp)        {            tmp/=10;            dig++;        }        int flag=0;        for(int i=n-dig*9;i<n;i++)        {            if(i+solve(i)==n)            {                flag=1;                cout<<i<<endl;                break;            }        }        if(!flag)            cout<<0<<endl;    }    return 0;}

解答：

紫書上的例題，蠻簡單的，書上是打表做的。

其實可以考慮一下，一個數的產生元不可能超過它本身，那最小的時候也就是n減去每一位全都是9的情況。
如此下來枚舉即可