Digit Generator

最新推荐文章于 2023-12-03 18:04:03 发布

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本文提供了一个算法，旨在找到给定整数的最小生成数，即该整数加上其各位数字之和等于自身的数字。通过实例演示和代码实现，深入解析数字与生成数之间的关系。

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Description
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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 $\le$ N $\le$ 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
Sample Output
198
0
1979
就是一个函数，函数值是这个数加上每个位的数字的和。现在给你数字，求函数值为这个数字最小的数。
直接开叔祖标记一下，从前向后跑一边就行了，之后的就是查询了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
int dp[100005];
int main()
{
    int i, j, m, n, ans, sum, temp;
    memset(dp, 0, sizeof(dp));
    for (i = 1; i <= 100000; i++)
    {
        sum = 0;
        temp = i;
        sum += temp;
        while (temp != 0)
        {
            sum += temp % 10;
            temp = temp / 10;
        }
        if (sum <= 100000)
        {
            if (dp[sum] == 0)
                dp[sum] = i;
        }
    }
    cin >> n;
    for (i = 1; i <= n; i++)
    {
        cin >> temp;
        cout << dp[temp] << endl;
    }
    return 0;
}