Digit Generator UVA - 1583（生成元）_1583s778o2r-优快云博客

本文介绍了一个算法问题，即找出一个给定整数的最小生成器。对于正整数N，其数字和定义为N本身与其各位数字之和。如果M是N的数字和，则称N为M的一个生成器。文章提供了C++实现代码，通过预先计算所有可能的生成器并存储在一个数组中，从而快速响应查询。

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For a positive integer N , the digit-sum of N is dened 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  nd 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  rst line of the input. Each test case takes one line containing an integer N , 1 <= N <= 100000.
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
197

直接打表。。。。。

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<iomanip>

using namespace std;

const int maxn = 1e5+7;

int a[maxn];

int getnum(int n){
    int sum = n;
    while(n){
        sum += (n%10);
        n /= 10;
    }
    return sum;
}

void init()
{
    memset(a,255,sizeof(a));
    for(int i = 0; i < maxn; i++)
    {
        int x = getnum(i);
        if(a[x] == -1) a[x] = i;
    }
    for(int i = 0; i < maxn; i++)
    {
        if(a[i] == -1) a[i] = 0;
    }
}

int main()
{
    int t,n;
    scanf("%d",&t);
    //cout << a[0] << endl;
    init();
    while(t--)
    {
            scanf("%d",&n);
            printf("%d\n",a[n]);
    }
    return 0;
}