题意:
解法:
满足条件的前缀:
1.前面一半要等于后面一半.
2.前面一半和后面一半都是回文.
判断条件1可以用字符串hash,
判断条件2也能用字符串hash.
因此hash一下就行了。
code:
#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
using namespace std;
const int maxm=2e5+5;
const int p=1331;
ull h[maxm];
ull h2[maxm];
ull base[maxm];
char s[maxm];
int n;
ull ask(int l,int r){
return h[r]-base[r-l+1]*h[l-1];
}
ull ask2(int l,int r){
return h2[l]-base[r-l+1]*h2[r+1];
}
bool is(int l,int r){
if(l==r)return 1;
int mid=(l+r)/2;
if((r-l+1)%2==0){
return ask(l,mid)==ask2(mid+1,r);
}else{
return ask(l,mid-1)==ask2(mid+1,r);
}
}
void solve(){
cin>>(s+1);
n=strlen(s+1);
base[0]=1;
for(int i=1;i<maxm;i++){
base[i]=base[i-1]*p;
}
for(int i=1;i<=n;i++){
h[i]=h[i-1]*p+s[i];
}
for(int i=n;i>=1;i--){
h2[i]=h2[i+1]*p+s[i];
}
int ans=0;
for(int i=1;i<=n;i++){
int mid=(1+i)/2;
if(i%2){
if(!is(1,mid)||!is(mid,i))continue;
if(ask(1,mid)==ask(mid,i))ans++;
}else{
if(!is(1,mid)||!is(mid+1,i))continue;
if(ask(1,mid)==ask(mid+1,i))ans++;
}
}
cout<<ans<<endl;
}
signed main(){
ios::sync_with_stdio(0);
solve();
return 0;
}