BZOJ 3864: Hero meet devil

双倍经验题

同BZOJ5336

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod=1e9+7;
int n,K,now[25],G[25],To[50005][4],F[2][50005],ED[50005],ANS[25],S[25];
char s[1000005];
void dfs(int t,int s){
	if (t>K){
		for (int i=1; i<=K; i++) {
			now[i]=now[i-1];
			if (s&(1<<i-1)) now[i]++;
		}
		ED[s]=now[K];
		for (int to=0; to<4; to++){
			for (int i=1; i<=K; i++) G[i]=max(max(G[i-1],now[i]),now[i-1]+(to==S[i]));
			for (int i=K; i>=1; i--) (To[s][to]<<=1)|=(G[i]-G[i-1]);
		}
		return;
	}
	dfs(t+1,s<<1);
	dfs(t+1,s<<1|1);
}
int main(){
	int T;
	scanf("%d",&T);
	while (T--){
		scanf("%s",s+1);
		K=strlen(s+1);
		scanf("%d",&n);
		for (int i=1; i<=K; i++){
			if (s[i]=='A') S[i]=0;
			else if (s[i]=='C') S[i]=1;
			else if (s[i]=='G') S[i]=2;
			else S[i]=3;
		}
		memset(To,0,sizeof(To));
		dfs(1,0);
		memset(F,0,sizeof(F));
		F[0][0]=1;
		for (int i=0; i<n; i++){
			for (int pre=0; pre<(1<<K); pre++) F[(i+1)%2][pre]=0;
			for (int pre=0; pre<(1<<K); pre++)
				for (int to=0; to<4; to++)
					(F[(i+1)%2][To[pre][to]]+=F[i%2][pre])%=mod;
		}
		memset(ANS,0,sizeof(ANS));
		for (int now=0; now<(1<<K); now++) (ANS[ED[now]]+=F[n%2][now])%=mod;
		for (int i=0; i<=K; i++) printf("%d\n",ANS[i]);
	}
	return 0;
}

  

posted @ 2018-10-19 15:52  ~Silent  阅读(130)  评论(0编辑  收藏  举报
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