UVA 10396 - Vampire Numbers(暴力打表)

Problem D

Vampire Numbers

Input: standard input

Output: standard output

Time Limit: 5 seconds

 

A number v = xy with an even number (n) of digits formed by multiplying a pair of n/2-digit numbers (where the digits are taken from the original number in any order) x and y together is known as vampire number. Pairs of trailing zeros (Both the numbers have a trailing zero) are not allowed. If v is a vampire number then x and y are called its "fangs." Examples of 4-digit vampire numbers include

 

1) 21 x 60 = 1260
2) 15 x 93 = 1395
3) 35 x 41 = 1435
4) 30 x 51 = 1530
5) 21 x 87 = 1827
6) 27 x 81 = 2187
7) 80 x 86 = 6880

 

In this program you will have to find all the 4, 6 and 8 digit even vampire numbers.

 

Input

The input file contains maximum ten lines of input. Each line contains a single integer n whose value is 4, 6 or 8. Input is terminated by end of file.

 

Output

For each input n produce all the n-digit vampire numbers that are even in ascending order. Print a blank line after the output for each set of input.

 

Sample Input:

4

4

 

Sample Output:

1260

1530

6880

 

1260

1530

6880

题意：输出所有吸血鬼偶数，满足条件为分成n/2的两位数相乘等于本身,并且两个数不能同时尾随0.

思路：暴力枚举，每一个都去判断，如果是两个奇数相乘必然为奇数直接continue，答案全放入set里面，自动排序去重，最后输出，这样跑了7秒。但是n其实只有4，6，8.每一次求出来的答案保留下来，下次继续用，优化到2秒多。

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <set>using namespace std;const int ten[5] = {1, 10, 100, 1000, 10000};int n, vis[10], v[10];set<int> ans[10];bool judge(int x, int y) {memset(vis, 0, sizeof(vis));int num = x * y;while (x) {vis[x % 10]++;x /= 10;}while (y) {vis[y % 10]++;y /= 10;}while (num) {vis[num % 10]--;num /= 10;}for (int i = 0; i < 10; i++)if (vis[i]) return false;return true;}void solve() {if (!v[n]) { for (int i = ten[n / 2 - 1]; i < ten[n / 2]; i++)for (int j = i; j < ten[n / 2]; j++) {if (i % 2 && j % 2) continue;if (!(i % 10) && !(j % 10)) continue;if (judge(i, j))ans[n].insert(i * j);}}v[n] = 1;for (set<int>::iterator it = ans[n].begin(); it != ans[n].end(); it++)printf("%d\n", *it);printf("\n");}int main() {while (~scanf("%d", &n)) {solve();}return 0;}