lightoj 1021 - Painful Bases（数位dp+状压）

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1021

 

题解：简单的数位dp由于总共就只有16个存储一下状态就行了。求各种进制能否整除k模仿10进制就行。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
ll dp[1 << 17][21] , po[20][20];
int base , k , num[20];
char s[20];
ll dfs(int len , int stat , int mod , int Max) {
    if(len == 0) {
        if(mod % k == 0) return 1;
        return 0;
    }
    if(dp[stat][mod] != -1) return dp[stat][mod];
    ll sum = 0;
    for(int i = 0 ; i < Max ; i++) {
        if((1 << i) & stat) continue;
        else {
            sum += dfs(len - 1 , stat | (1 << i) , (num[i] + mod * base) % k, Max);
        }
    }
    dp[stat][mod] = sum;
    return sum;
}
int main() {
    int t;
    scanf("%d" , &t);
    int ans = 0;
    for(int i = 2 ; i <= 16 ; i++) {
        po[i][0] = 1;
        for(int j = 1 ; j <= 16 ; j++) {
            po[i][j] = po[i][j - 1] * i;
        }
    }
    while(t--) {
        scanf("%d%d" , &base , &k);
        scanf("%s" , s);
        int len = strlen(s);
        for(int i = 0 ; i < len ; i++) {
            if(s[i] == 'A') num[i] = 10;
            else if(s[i] == 'B') num[i] = 11;
            else if(s[i] == 'C') num[i] = 12;
            else if(s[i] == 'D') num[i] = 13;
            else if(s[i] == 'E') num[i] = 14;
            else if(s[i] == 'F') num[i] = 15;
            else num[i] = s[i] - '0';
        }
        for(int i = 0 ; i < (1 << len) ; i++) {
            for(int j = 0 ; j <= k ; j++) dp[i][j] = -1;
        }
        printf("Case %d: %lld\n" , ++ans , dfs(len , 0 , 0 , len));
    }
    return 0;
}