BZOJ 2084: [Poi2010]Antisymmetry(Hash+二分)

求一个01序列的子串取反并反转后与原串相同的个数.

很显而易见的是，反转的话只要子串对应的i和n-i+1位相反即可,这个看一下样例能很快看出来.

所以我们正着求一遍hash,反着取反然后求hash.

枚举中间点,二分一下这个子串长度的一半,check的话就是判断前一半子串的正hash值与后一半子串取反后的反hash值是否相等.

答案怎么统计呢,显然,如果一个子串满足条件,那么去掉首位得到的子串也满足条件,ans+=len/2...

然而细节写错wa了2发。。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#define ll long long
#define out(a) printf("%lld ",a)
#define writeln printf("
")
const int N=5e5+50;
const int MOD=1e9+7;
const int base=233;
using namespace std;
int n;
char s[N];
ll Hash1[N],Hash2[N],Pow[N];
ll ans;
int read()
{
     int s=0,t=1; char c;
     while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
     while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
     return s*t;
}
ll readl()
{
     ll s=0,t=1; char c;
     while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();}
     while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();}
     return s*t;
}
ll get(int l,int r,int x)
{
    if (x==1) return (Hash1[r]-(ll)Hash1[l-1]*Pow[r-l+1]%MOD+MOD)%MOD;
    return (Hash2[l]-(ll)Hash2[r+1]*Pow[r-l+1]%MOD+MOD)%MOD;
}
int solve(int x)
{
    int l=1,r=min(x,n-x),mid=0; bool flag=false;
    while (l<=r){
      mid=(l+r)>>1;
      if (get(x-mid,x+mid-1,1)==get(x-mid,x+mid-1,2)) flag=true,l=mid+1;
      else r=mid-1;
    }
    //out(x),out(r),writeln;
    if (flag) return r;
    return 0;
}
int main()
{
    n=read(); Pow[0]=1;
    scanf("%s",s);
    for (int i=1;i<=n;i++)
      Pow[i]=Pow[i-1]*base%MOD;
    for (int i=0;i<n;i++)
      Hash1[i]=(Hash1[i-1]*base+s[i])%MOD;
    for (int i=n-1;i>=0;i--)
      Hash2[i]=(Hash2[i+1]*base+(48+49-s[i]))%MOD;
    for (int i=1;i<n;i++)
      ans+=solve(i);
    out(ans);
    return 0;
}