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  • 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
    ",ANS[i]);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/silenty/p/9816921.html
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