[CLYZ2017]day18

子集

solution

60分

枚举中位数,枚举集合大小,贪心使平均值尽量大,取最大值.

#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 2005
using namespace std;
double a[N],ans,sum;
int n;
inline void Aireen(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%lf",&a[i]);
	sort(a+1,a+1+n);
	for(int i=2,k;i<n;++i){
		sum=a[i];k=1;
		for(int j=i-1,l=n;j&&l>i;--j,--l){
			sum+=a[j]+a[l];k+=2; 
			ans=max(ans,sum/k-a[i]);
		}
	}
	for(int i=3,k;i<n;++i){
		sum=a[i-1]+a[i];k=2;
		for(int j=i-2,l=n;j&&l>i;--j,--l){
			sum+=a[j]+a[l];k+=2; 
			ans=max(ans,sum/k-(a[i-1]+a[i])/2.0);
		}
	}
	printf("%.5lf\n",ans);
}
int main(){
	freopen("subset.in","r",stdin);
	freopen("subset.out","w",stdout);
	Aireen();
	fclose(stdin);
	fclose(stdout);
	return 0;
}

100分

\(yy\)一下发现中位数为一个数时值更大.
枚举中位数,可以发现这是一个关于集合大小的单峰函数,三分即可.

#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 200005
using namespace std;
int n,l,r,m1,m2;
double a[N],s[N],ans;
inline double f(int x,int i){
	return (s[n]-s[n-x]+s[i]-s[i-x-1])/(x<<1|1)-a[i];
}
inline void Aireen(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%lf",&a[i]);
	sort(a+1,a+1+n);
	for(int i=1;i<=n;++i) 
		s[i]=s[i-1]+a[i];
	for(int i=1;i<=n;++i){
		l=1;r=min(n-i,i-1);
		while(l+2<r){
			m1=l+(r-l)/3;m2=r-(r-l)/3;
			if(f(m1,i)<f(m2,i)) l=m1;
			else r=m2;
		}
		for(int j=l;j<=r;++j)
			ans=max(ans,f(j,i));
	}
	printf("%.5lf\n",ans);
}
int main(){
	freopen("subset.in","r",stdin);
	freopen("subset.out","w",stdout);
	Aireen();
	fclose(stdin);
	fclose(stdout);
	return 0;
}

子串

100分

将字符串建成\(AC\)自动机,\(f[i]\)表示从节点\(i\)到结束的期望步数.
\(f[i]=1+\frac{1}{26}\sum_{j='a'}^{'z'}f[tr[i][j]]\).
如果\(i\)为结束节点,则\(f[i]=0\).
根据关系式高斯消元即可(分母移项即为整式方程).

#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define K 26
#define M 15
#define N 155
#define P 1000000007
using namespace std;
typedef long long ll;
int a[N][N],tr[N][K],nxt[N],f[N],n,t,cnt;
char c[M];
queue<int> q;
inline void insert(){
	int u=0,l=strlen(c+1);
	for(int i=1;i<=l;++i){
		c[i]-='a';
		if(!tr[u][c[i]])
			tr[u][c[i]]=++cnt;
		u=tr[u][c[i]];
	}
	f[u]=1; 
}
inline void bfs(int u){
	for(int i=0;i<K;++i)
		if(tr[u][i]){
			nxt[tr[u][i]]=0;
			q.push(tr[u][i]);
		}
	while(!q.empty()){
		u=q.front();q.pop();
		for(int i=0,j;i<K;++i)
			if(tr[u][i]){
				q.push(tr[u][i]);j=nxt[u];
				while(j&&!tr[j][i]) j=nxt[j];
				nxt[tr[u][i]]=tr[j][i];
			}
			else{
				j=nxt[u];
				while(j&&!tr[j][i]) j=nxt[j];
				tr[u][i]=tr[j][i];
			}
	}
}
inline int rev(int x){
	int ret=1,k=P-2;
	while(k){
		if(k&1) ret=1ll*ret*x%P;
		x=1ll*x*x%P;k>>=1;
	}
	return ret;
}
inline int gauss(int n){
	int tmp;
	for(int i=0;i<=n;++i){
		if(!a[i][i]) 
			for(int j=i+1;j<=n;++j)
				if(a[j][i]){
					for(int k=i;k<=n+1;++k)
						swap(a[i][k],a[j][k]);
					break;
				}
		for(int j=0;j<=n;++j)
			if(j!=i&&a[j][i]){
				tmp=1ll*a[j][i]*rev(a[i][i])%P;
				for(int k=i;k<=n+1;++k)
					a[j][k]=(a[j][k]-1ll*tmp*a[i][k]%P)%P;
			}
	}
	for(int i=0;i<=n;++i)
		f[i]=1ll*a[i][n+1]*rev(a[i][i])%P;
	if(f[0]<0) f[0]+=P;
	return f[0];
}
inline void Aireen(){
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n); 
		cnt=0;
		memset(f,0,sizeof(f));
		memset(tr,0,sizeof(tr));
		for(int i=1;i<=n;++i){
			scanf("%s",c+1);insert();
		}
		bfs(0);
		memset(a,0,sizeof(a));
		for(int i=0;i<=cnt;++i)
			if(f[i]){
				a[i][i]=1;a[i][cnt+1]=0;
			}
			else{
				a[i][i]=a[i][cnt+1]=26;
				for(int j=0;j<K;++j)
					--a[i][tr[i][j]];
			}
		printf("%d\n",gauss(cnt));
	}
}
int main(){
	freopen("substring.in","r",stdin);
	freopen("substring.out","w",stdout);
	Aireen();
	fclose(stdin);
	fclose(stdout);
	return 0;
}