Problem 1068 An Interesting Set

Accept: 641 Submit: 1172 Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

Thomas, a mathematician, is interested in a set S consisting of positive integers. Set S is defined as follows:

(1) If a positive integer satisfies the property that the sum of all digits of it is equal to the product of them, then it belongs to S.

(2) If m belongs to S, then 2m belongs to S. Thomas wants to find out the Kth smallest integer in set S. Design a program to solve the problem.

Input

The first line in the input is an integer N representing the number of test cases. The following N lines are N test cases and each line contains an integer K (1<=K<=500).

Output

For each test case, output a line containing an integer that is the Kth smallest integer in set S.

Sample Input

2

5

16

Sample Output

5

22

Source

FJNUPC 2005

解决方案

使用打表的方法

让i从1开始判断，直到得到500个符合条件的数值，存入数组S

判断的条件有两个，一是判断i是否是S中原有数值的2倍，二是判断i的各位数字的总和是否和乘积相等

判断条件一可以利用set迅速查找指定key值的特性

判断条件二利用取余即可

int num[505]; set<int> s; set<int>::iterator it; bool isin(int a) { int total = 0, muti = 1; int num; //条件一，需要判断是否偶数 if (a % 2 == 0) { //否则 11 / 2 = 5，即所有数字都会被认为符合条件一 it = s.find(a / 2); if (it != s.end()) { return true; } } //条件二 while (a) { num = a % 10; if (num == 0) { return false; } total += num; muti *= num; a /= 10; } if (total != muti) { return false; } return true; } //打表初始化 void init() { int i, j; for (i = 1, j = 1; j <= 500; i++) { if (isin(i)) { num[j++] = i; s.insert(set<int>::value_type(i)); } } } int main() { int N, k; cin >> N; init(); while (N--) { cin >> k; cout << num[k] << endl; } }