SGU 310. Hippopotamus( 状压dp )

题目大意:给N块板, 有A,B2种类型的板, 要求任意M块连续的板中至少有K块B板.1≤n≤60,1≤m≤15,0≤k≤m≤n.

dp(x, s)表示第x块板, x前M块板的状态为s, 然后合法状态转移就行了. 

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#include<cstdio>

#include<cstring>

#include<algorithm>

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 61;

const int maxm = 21;

 

#define b(x) (1 << (x))

 

int N, M, K, cnt[b(maxm)];

ll dp[2][b(maxm)];

 

int main() {

scanf("%d%d%d", &N, &M, &K);

int c = 0, p = 1;

for(int s = b(M); s--; ) {

cnt[s] = 0;

for(int i = 0; i < M; i++)

if(s & b(i)) cnt[s]++;

dp[c][s] = (cnt[s] >= K);

}

int t = b(M - 1);

for(int i = N - M; i--; ) {

swap(c, p);

memset(dp[c], 0, sizeof dp[c]);

for(int s = b(M); s--; ) if(cnt[s >> 1] >= K) {

dp[c][s >> 1] += dp[p][s];

dp[c][s >> 1 | t] += dp[p][s];

} else if(cnt[s >> 1] + 1 == K)

dp[c][s >> 1 | t] += dp[p][s];

}

ll ans = 0;

for(int i = b(M); i--; )

if(cnt[i] >= K) ans += dp[c][i];

printf("%lld\n", ans);

return 0;

}

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310. Hippopotamus

Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

After fixing your roof, you still think that it looks unpretty. So you opt for a new one, consisting of n consecutive long narrow boards. You have two types of boards: wooden ones and iron ones, giving you an amazing total of 2ⁿ possible roofs.

But the safety should not be left aside. Having considered the weight and the cruising speed of a falling hippopotamus, you decide to have at least k iron boards among every m consecutive boards.

How many possibilities do you have?

Input

The input file contains three integers, n, m and k, separated by spaces and/or line breaks. 1 ≤ n ≤ 60, 1 ≤ m ≤ 15, 0 ≤ k ≤ m ≤ n.

Output

Output the number of possibilities.

Example(s)

sample input	sample output
10 2 1	144

sample input	sample output
5 5 2	26

sample input	sample output
3 2 2	1