Even NOI Day1T1 will not do. . . Read the solution to a problem can not write to Claris also copied code. .
Topic links: (luogu) https://www.luogu.org/problemnew/show/P1117
(bzoj)https://www.lydsy.com/JudgeOnline/problem.php?id=4650
(uoj) http://uoj.ac/problem/219
answer:
\ (f [i] \) represented by \ (I \) end \ (AA \) the number of sub-string, \ (G [I] \) represented by \ (I \) starting \ (AA \) the number of sub-type string
How demand \ (f, G \) ?
Break the mindset, who said a request it must be a
Sublength to seek
Enumeration length \ (L \) , handles all of the length \ (2L \) a \ (AA \) type of the substring \ (F \) and \ (G \) contributed
If every \ (L \) to a discharge point timer length, Bah, key
Then any \ (AA \) type substring will go through two adjacent key points
First, we must meet the characters on these two key positions as
On this basis, the maximum number is determined to move forward in the future the same
This would translate into LCP and LCS problem, and the two adjacent key points \ (f, g \) the impact of the array is the interval +1 prefix and solve differential
Then push a push on the line, do not push the wrong note + 1-1
The time complexity is the harmonic series, \ (O (n-\ log n-) \)
Code
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define llong long long
using namespace std;
const int N = 1<<15;
const int LGN = 15;
const int S = 26;
int log2[N+3];
struct SparseTable
{
int n;
int str[N+3];
int rk[N+3];
int tmp[N+3];
int height[N+3];
int h[N+3];
int sa[N+3];
int wb[N+3];
int mini[N+3][LGN+3];
void get_sa()
{
int *x = rk,*y = tmp;
for(int i=0; i<=S; i++) wb[i] = 0;
for(int i=1; i<=n; i++) wb[x[i]=str[i]]++;
for(int i=1; i<=S; i++) wb[i] += wb[i-1];
for(int i=n; i>=1; i--) sa[wb[x[i]]--] = i;
int s = S,p = 0;
for(int j=1; p<n; j<<=1)
{
p = 0;
for(int i=n-j+1; i<=n; i++) y[++p] = i;
for(int i=1; i<=n; i++) if(sa[i]>j) y[++p] = sa[i]-j;
for(int i=1; i<=s; i++) wb[i] = 0;
for(int i=1; i<=n; i++) wb[x[y[i]]]++;
for(int i=1; i<=s; i++) wb[i] += wb[i-1];
for(int i=n; i>=1; i--) sa[wb[x[y[i]]]--] = y[i];
swap(x,y);
p = 1; x[sa[1]] = 1;
for(int i=2; i<=n; i++) x[sa[i]] = (y[sa[i]]==y[sa[i-1]] && y[sa[i]+j]==y[sa[i-1]+j]) ? p : ++p;
s = p;
}
for(int i=1; i<=n; i++) rk[sa[i]] = i;
for(int i=1; i<=n; i++)
{
h[i] = h[i-1]==0 ? 0 : h[i-1]-1;
while(i+h[i]<=n && sa[rk[i-1]]+h[i]<=n && str[i+h[i]]==str[sa[rk[i]-1]+h[i]])
{
h[i]++;
}
}
for(int i=1; i<=n; i++) height[i] = h[sa[i]];
for(int i=1; i<=n; i++) mini[i][0] = height[i];
for(int j=1; j<=LGN; j++)
{
for(int i=1; i+(1<<j)-1<=n; i++)
{
mini[i][j] = min(mini[i][j-1],mini[i+(1<<j-1)][j-1]);
}
}
}
int querymin(int lb,int rb)
{
int g = log2[rb-lb+1];
return min(mini[lb][g],mini[rb-(1<<g)+1][g]);
}
int LCP(int x,int y)
{
if(x==y) return n-x+1;
if(rk[x]>rk[y]) swap(x,y);
return querymin(rk[x]+1,rk[y]);
}
void clear()
{
for(int i=1; i<=n; i++) str[i] = rk[i] = tmp[i] = height[i] = h[i] = sa[i] = wb[i] = 0;
for(int i=1; i<=n; i++)
{
for(int j=0; j<=LGN; j++)
{
mini[i][j] = 0;
}
}
}
} s1,s2;
llong f[N+3];
llong g[N+3];
char a[N+3];
int n;
int LCP(int x,int y) {return s1.LCP(x,y);}
int LCS(int x,int y) {return s2.LCP(n+1-x,n+1-y);}
void preprocess()
{
log2[1] = 0; for(int i=2; i<=N; i++) log2[i] = log2[i>>1]+1;
}
void clear()
{
s1.clear(); s2.clear();
for(int i=0; i<=n+1; i++) a[i] = 0,f[i] = g[i] = 0ll;
}
int main()
{
preprocess();
int T; scanf("%d",&T);
while(T--)
{
scanf("%s",a+1); n = strlen(a+1); for(int i=1; i<=n; i++) a[i]-=96;
for(int i=1; i<=n; i++) s1.str[i] = a[i]; s1.n = n;
s1.get_sa();
for(int i=1; i<=n; i++) s2.str[i] = a[n+1-i]; s2.n = n;
s2.get_sa();
for(int i=1; i+i<=n; i++)
{
for(int j=i+i; j<=n; j+=i)
{
if(a[j]==a[j-i])
{
int lb = j-LCS(j,j-i)+1,rb = j+LCP(j,j-i)-1;
lb = max(lb+i-1,j); rb = min(rb,j+i-1);
if(lb<=rb)
{
f[lb]++; f[rb+1]--;
g[lb-i-i+1]++; g[rb+1-i-i+1]--;
}
}
}
}
for(int i=1; i<=n; i++) f[i] += f[i-1],g[i] += g[i-1];
llong ans = 0ll;
for(int i=1; i<n; i++)
{
llong tmp = f[i]*g[i+1];
ans += tmp;
}
printf("%lld\n",ans);
clear();
}
return 0;
}