题目链接：uva 10825 - Anagram and Multiplication

题目大意：给出m和n，要求找一个m位的n进制数，要求说该数乘以2~m中的任意一个数的结果是原先数各个位上数值的一个排序。

解题思路：枚举最后一位数，然后用这个数去乘以2~m并对n取模，然后得到的数一定就是这个数的组成，暴力搜索一下并判断。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 405;
const int maxm = 10;

int m, n, a[maxn], v[maxn], ans[maxm];

bool check (int x, int* b) {
    memset(a, 0, sizeof(a));
    int tmp = 0;

    for (int i = m-1; i >= 0; i--) {
        tmp = tmp + x * b[i];

        int k = tmp % n;
        a[k]++;
        tmp = tmp / n;

        if (a[k] > v[k])
            return false;
    }

    return tmp == 0;
}

bool judge (int x) {
    int tmp = 0;
    memset(v, 0, sizeof(v));

    for (int i = 0; i < m; i++) {
        tmp = (x + tmp) % n;
        ans[i] = tmp;
        v[ans[i]]++;
    }

    swap(ans[0], ans[m-1]);

    sort(ans, ans + m - 1);
    do {
        bool flag = true;
        if (ans[0] == 0)
            continue;
        for (int i = 2; i <= m; i++) {
            if (check(i, ans)) 
                continue;
            flag = false;
            break;
        }

        if (flag)
            return true;

    } while (next_permutation(ans, ans + m - 1));

    return false;
}

int main () {
    while (scanf("%d%d", &m, &n) == 2 && n + m) {
        bool flag = true;
        for (int i = 1; i < n; i++) {
            if (judge(i)) {
                flag = false;
                break;
            }
        }

        if (flag)
            printf("Not found.\n");
        else {
            printf("%d", ans[0]);
            for (int i = 1; i < m; i++)
                printf(" %d", ans[i]);
            printf("\n");
        }
    }
    return 0;
}