uva 10396 - Vampire Numbers(暴力)

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题目链接：uva 10396 - Vampire Numbers

题目大意:输入n，找出为数为n的吸血鬼数字。吸血鬼数字即偶数，并且可以拆分成两个不同为10的倍数的数，相乘为它本身。

解题思路：枚举拆分后的两个数，然后判断。

#include <stdio.h>#include <string.h>#include <set>#include <algorithm>using namespace std;const int N = 6;const int tmp[N] = {0, 1, 10, 100, 1000, 10000};int n, rec[N];set<int> v[N];bool judge(int a, int b) {int c[10], s = a * b;memset(c, 0, sizeof(c));while (a) {c[a%10]++; a /= 10;}while (b) {c[b%10]++; b /= 10;}while (s) {c[s%10]--; s /= 10;}for (int i = 0; i < 10; i++) if (c[i] != 0) return false;return true;}void solve() {int k = n / 2;if (rec[k] == 0) {for (int i = tmp[k]; i < tmp[k+1]; i++) {for (int j = i; j < tmp[k+1]; j++) {if (i % 10 == 0 && j % 10 == 0) continue;if (i % 2 && j % 2) continue;if (judge(i, j)) v[k].insert(i * j);}}}rec[k] = 1;for (set<int>::iterator i = v[k].begin(); i != v[k].end(); i++)printf("%d\n", *i);printf("\n");}int main() {memset(rec, 0, sizeof(rec));while (scanf("%d", &n) == 1) {solve();}return 0;}

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