UVa 10396 - Vampire Numbers (暴力)

题意

计算n位的吸血鬼数有几个。

满足以下条件的成为吸血鬼数。

1. v = xy, xy中的数合起来正好和v中的一样。
2. xy不能同时有后缀0

思路

一开始打了个表，不过太大了，不让交。然后就直接暴力了。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
bool Check(int a, int b)
{
    int vis[11], vvis[11];
    MS(vis, 0); MS(vvis, 0);
    int num = a * b;
    while (a)
    {
        int rem = a % 10;
        a /= 10;
        vis[rem]++;
    }
    while (b)
    {
        int rem = b % 10;
        b /= 10;
        vis[rem]++;
    }
    while (num)
    {
        int rem = num % 10;
        num /= 10;
        vvis[rem]++;
    }
    for (int i = 0; i < 10; i++)
        if (vis[i] != vvis[i]) return false;
    return true;
}
set<int> ans[10];
int main()
{
    const LL MAX = 1e10;
    int i, j;
    for (i = 10; i <= 99; i++)
        for (j = i; j <= 99; j++)
        {
            int num = i * j;
            if (num & 1) continue;
            if (i % 10 == 0 && j % 10 == 0) continue;
            if (Check(i, j)) ans[2].insert(num);
        }
    for (i = 100; i <= 999; i++)
        for (j = 100; j <= 999; j++)
        {
            int num = i * j;
            if (num > 100000000) continue;
            if (num & 1) continue;
            if (i % 10 == 0 && j % 10 == 0) continue;
            if (Check(i, j)) ans[3].insert(num);
        }
    for (i = 1000; i <= 9999; i++)
        for (j = i; j <= 9999; j++)
        {
            if (i % 10 == 0 && j % 10 == 0) continue;
            LL num = (LL)i * j;
            if (num > MAX) continue;
            if (num & 1) continue;
            if (Check(i, j)) ans[4].insert(num);
        }
    int n;
    while (~scanf("%d", &n))
    {
        int pos = n / 2;
        for (set<int>::iterator it = ans[pos].begin(); it != ans[pos].end(); it++)
            printf("%d\n", *it);
        puts("");
    }
    return 0;
}