AM文字列
問題解決のアイデア：ハッシュ。プレフィックスの長さはすでにわかっているため、プレフィックスの対称ハッシュ値は、比較のためにハッシュによって簡単に計算できます。
バイナリ文字列のハッシュを統合します
| 0 | 1 | 2 | 3 | 4 | 5 | 5 | |
|---|---|---|---|---|---|---|---|
| ビット | 1 | 2 | 4 | 8 | 16 | 32 | 64 |
| バイナリ | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
| a | 1 | 2 | 5 | 10 | 21 | 43 | 86 |
2〜4の2進数「101」を見つけます。a[4] -a [1] * bit [3]、
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double lf;
typedef unsigned long long ull;
typedef pair<ll,int>P;
const int inf = 0x7f7f7f7f;
const ll INF = 1e16;
const int N = 1e6+10;
const int M = 50005;
const ll base = 131;
const ll mod = 1000000007;
const double PI = acos(-1.0);
const double eps = 1e-4;
inline int read(){
int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=x*10+ch-'0';ch=getchar();}return x*f;}
inline string readstring(){
string str;char s=getchar();while(s==' '||s=='\n'||s=='\r'){
s=getchar();}while(s!=' '&&s!='\n'&&s!='\r'){
str+=s;s=getchar();}return str;}
int random(int n){
return (int)(rand()*rand())%n;}
void writestring(string s){
int n = s.size();for(int i = 0;i < n;i++){
printf("%c",s[i]);}}
ll fast_power(ll a,ll p){
ll ret = 1;
while(p){
if(p&1) ret = (ret*a)%mod;
p >>= 1;
a = (a*a)%mod;
}
return ret;
}
char str[N];
ll a[N],b[N];
ll bit[N];
int main(){
//srand((unsigned)time(NULL));
//freopen( "out.txt","w",stdout);
scanf("%s",str+1);
bit[0] = 1;
for(int i = 1;i < N;i++){
bit[i] = (bit[i-1]*base);
}
int n = strlen(str+1);
for(int i = 1;i <= n;i++){
a[i] = a[i-1]*base+(str[i]-97);
}
for(int i = n;i >= 1;i--){
b[i] = b[i+1]*base+(str[i]-97);
}
int ans = 0;
for(int i = 1;i <= n;i++){
if(i+i <= n){
ll x = a[i];
ll p = b[1]-b[i+1]*bit[i];
ll y = a[i+i]-a[i]*bit[i];
ll t = b[i+1]-b[i+i+1]*bit[i];
if(x==y&&x==t&&x==p) ans++;
}
if(i+i-1 <= n){
ll x = a[i];
ll y = a[i+i-1]-a[i-1]*bit[i];
ll t = b[i]-b[i+i]*bit[i];
if(x == y && y == t) ans++;
}
}
cout<<ans<<endl;
return 0;
}