lightoj 1021 - Painful Bases 【数位dp 状压】
题目链接：lightoj 1021 - Painful Bases
1021 - Painful Bases
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
As you know that sometimes base conversion is a painful task. But still there are interesting facts in bases.
For convenience let’s assume that we are dealing with the bases from 2 to 16. The valid symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. And you can assume that all the numbers given in this problem are valid. For example 67AB is not a valid number of base 11, since the allowed digits for base 11 are 0 to A.
Now in this problem you are given a base, an integer K and a valid number in the base which contains distinct digits. You have to find the number of permutations of the given number which are divisible by K. K is given in decimal.
For this problem, you can assume that numbers with leading zeroes are allowed. So, 096 is a valid integer.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a blank line. After that there will be two integers, base (2 ≤ base ≤ 16) and K (1 ≤ K ≤ 20). The next line contains a valid integer in that base which contains distinct digits, that means in that number no digit occurs more than once.
Output
For each case, print the case number and the desired result.
Sample Input
Output for Sample Input
3
2 2
10
10 2
5681
16 1
ABCDEF0123456789
Case 1: 1
Case 2: 12
Case 3: 20922789888000
PROBLEM SETTER: JANE ALAM JAN
题意：给定base进制的字符串，没有重复的字符。问你该串的全排列中有多少个数可以被十进制k整除。
思路：设置dp[s][i][j]为s状态下处理到第i位对kmod为j的方案数。发现第二维由第一维限制，去掉第二维，然后就可以过了。
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#define ll o<<1
#define rr o<<1|1
#define fi first
#define se second
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int MAXN = 3*1e5 + 10;
LL dp[1<<16][20];
int a[20];
int base, K, len;
LL DFS(int s, int pos, int yu) {
if(pos == -1) return yu == 0;
if(dp[s][yu] != -1) return dp[s][yu];
LL ans = 0;
for(int i = 0; i < len; i++) {
if(s & (1<<i)) continue;
ans += DFS(s + (1<<i), pos-1, (yu * base + a[i]) % K);
}
dp[s][yu] = ans;
return ans;
}
int V(char op) {
if(op >= '0' && op <= '9') return op - '0';
return op - 'A' + 10;
}
int main()
{
int t, kcase = 1; scanf("%d", &t);
while(t--) {
scanf("%d%d", &base, &K); char str[20];
scanf("%s", str); len = strlen(str);
for(int i = 0; i < len; i++) {
a[i] = V(str[i]);
}
for(int i = 0; i < (1<<len); i++) {
for(int j = 0; j < K; j++) {
dp[i][j] = -1;
}
}
printf("Case %d: %lld\n", kcase++, DFS(0, len-1, 0));
}
return 0;
}