[Bzoj1072] [SCOI2007] Perm arrangement (like pressure dp)

Topic links: https://www.lydsy.com/JudgeOnline/problem.php?id=1072

Shaped pressure dp, dp [i] [j] indicates the state is i, the d I j is taken as the number of programs, there dp [i | (1 << k)] [(j * 10 + a [k]) % d] + = dp [i] [j], where k is the number of i is not in the state.

At the same time, because there are duplicate numbers, so finally, get rid of duplicate numbers.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 3e5 + 10;
 5 const ll mod = 1e9 + 7;
 6 const ll INF = 1e18;
 7 int dp[1050][1010];
 8 int main() {
 9     int t;
10     scanf("%d", &t);
11     while (t--) {
12         int d;
13         char s[15];
14         int vis[15];
15         int a[15];
16         scanf("%s%d", s, &d);
17         int len = strlen(s);
18         memset(dp, 0, sizeof(dp));
19         memset(vis, 0, sizeof(vis));
20         for (int i = 0; i < len; i++)
21             a[i] = s[i] - '0', vis[a[i]]++;
22         dp[0][0] = 1;
23         for (int i = 0; i < (1 << len); i++) {
24             for (int j = 0; j < d; j++) {
25                 for (int k = 0; k < len; k++) {
26                     if ((i&(1 << k)) == 0)
27                         dp[i | (1 << k)][(j * 10 + a[k]) % d] += dp[i][j];
28                 }
29             }
30         }
31         int ans = dp[(1 << len) - 1][0];
32         for (int i = 0; i <= 9; i++)
33             for (int j = 1; j <= vis[i]; j++)
34                 ans /= j;
35         printf("%d\n", ans);
36     }
37     return 0;
38 }