Light OJ 1021 - Painful Bases(状态压缩DP)

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
const int INF = 1e9+7;
const int maxn = 1055;
LL dp[70000][25];///dp[状态][对k取余后的值] = 种类
int num[20];
char str[20];
int main()
{
int T, base, K, cas = 1;
scanf("%d", &T);

while(T --)
{
scanf("%d %d", &base, &K);
scanf("%s", str);
memset(dp, 0, sizeof(dp));
for(int i=0; str[i]; i++)
{
if(str[i] >= 'A' && str[i] <= 'F')
num[i] = str[i] - 'A' + 10;
else
num[i] = str[i] - '0';
}

int n = strlen(str);
int Lim = (1<<n) - 1;

dp[0][0] = 1;
for(int i=1; i<=Lim; i++)///每种状态
{
for(int j=0; j<n; j++)///判断这种状态的数字是否存在
{
if( (i&(1<<j)) )
{
int newSta = i - (1<<j);
for(int k=0; k<K; k++)
{
int p = (k*base + num[j])%K;
dp[i][p] += dp[newSta][k];
}
}
}
}
printf("Case %d: %lld\n",cas++, dp[Lim][0]);

}

return 0;
}