Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 30856 | Accepted: 10310 |
Description
Input
Output
Sample Input
30 25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18 82 78 74 70 66 67 64 60 65 80 0
Sample Output
5
Hint
题意:有N(1 <= N <=20000)个音符的序列来表示一首乐曲,每个音符都是1..88范围内的整数,现在要找一个重复的主题。“主题”是整个音符序列的一个子串,它需要满足如下条件:
# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
#define LL long long
#define N 20005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
using namespace std;
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int Rank[N],height[N],s[N],a[N],n;
char str1[N],str2[N];
int cmp(int *r,int a,int b,int k)
{
return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)
{
int i,j,p,*x=wa,*y=wb,*t;
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[x[i]=r[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[x[i]]]=i;
p=1;
j=1;
for(; p<n; j*=2,m=p)
{
for(p=0,i=n-j; i<n; i++) y[p++]=i;
for(i=0; i<n; i++) if(sa[i]>=j) y[p++]=sa[i]-j;
for(i=0; i<n; i++) wv[i]=x[y[i]];
for(i=0; i<m; i++) wsf[i]=0;
for(i=0; i<n; i++) wsf[wv[i]]++;
for(i=1; i<m; i++) wsf[i]+=wsf[i-1];
for(i=n-1; i>=0; i--) sa[--wsf[wv[i]]]=y[i];
t=x;
x=y;
y=t;
x[sa[0]]=0;
for(p=1,i=1; i<n; i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
}
}
void getheight(int *r,int n)
{
int i,j,k=0;
for(i=1; i<=n; i++) Rank[sa[i]]=i;
for(i=0; i<n; i++)
{
if(k)
k--;
else
k=0;
j=sa[Rank[i]-1];
while(r[i+k]==r[j+k])
k++;
height[Rank[i]]=k;
}
}
bool check(int x)
{
int imax, imin;
imax = imin = sa[1];
for(int i=2; i<=n; ++i)
{
if(height[i]>=x)
{
imax = max(imax, sa[i]);
imin = min(imin, sa[i]);
continue;
}
if(imax-imin>x) return true;
else imax = imin = sa[i];
}
return imax-imin>x;
}
int main()
{
while(~scanf("%d",&n),n)
{
for(int i=0; i<n; ++i) scanf("%d",&s[i]);
for(int i=0; i<n-1; ++i) s[i] = s[i+1]-s[i]+88;
s[--n] = 0;
getsa(s,sa,n+1,200);
getheight(s,n);
int L=0, R=(n+1)/2;
while(L<=R)
{
int mid = L+R>>1;
if(check(mid)) L = mid+1;
else R = mid-1;
}
++R;
printf("%d\n",R<5?0:R);
}
}