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LightOJ 1021 - Painful Bases (状压dp)

2015-11-02 17:59 344 查看
题意:

给你一个base进制的数,问这个数的全排列中,有多少个能被k整除

分析:

因为位数不多,所以可以状压

dp[s][m]:=状态为s,余数为m有多少个数,直接转移就好了

代码:

//
//  Created by TaoSama on 2015-11-01
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

typedef long long LL;
LL dp[1 << 16][20];
int n, b, k;
char a[20];

int get(char c) {
if(isdigit(c)) return c - '0';
return c - 'A' + 10;
}

LL dfs(int s, int m) {
if(s == (1 << n) - 1) return m == 0;
LL& ret = dp[s][m];
if(~ret) return ret;

ret = 0;
for(int i = 0; i < n; ++i) {
if(s >> i & 1) continue;
ret += dfs(s | 1 << i, (m * b + get(a[i])) % k);
}
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%d%d%s", &b, &k, a);
n = strlen(a);
memset(dp, -1, sizeof dp);
printf("Case %d: %lld\n", ++kase, dfs(0, 0));
}
return 0;
}
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标签:  dp lightoj