uva 10825 – Anagram and Multiplication(暴力)

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中的随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，暴力搜索一下并推断。

#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;
}