英文
Description
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
题目大意：
如果x加上x的各个数字之和得到y，就说x是y的生成元。给出n（1<=n<=100000），求最小生成元。无解输出0 。例如，n=216,121,2005时的解分别为198,0,1979 。
提示：
只要枚举就行了。但单纯的枚举容易炸，比如j要从n-45开始，不要从1开始，会快很多。f1是用来求数字和的。代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;
int T,n;
bool bo;
int f1(int k)
{
int ret=0;
while(k>0)
{
ret+=k%10;
k/=10;
}
return ret;
}
int main()
{
scanf("%d",&T);
for(int i=1;i<=T;i++)
{
bo=0;
scanf("%d",&n);
for(int j=n-45;j<=n;j++) if(f1(j)+j==n)
{
bo=1;
printf("%d\n",j);
break;
}
if(bo==0) printf("0\n");
}
return 0;
}相关链接：
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