未理解透，鬼知道怎么A的

蒟蒻交了个乱猜贪心搞了10pts，一翻题解群佬乱舞，最后DP解决

$\exists i - next[i] <= j, f[j] = f[next[i]] $

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 500007;

char str[N];
int nxt[N];
int f[N], pos[N];
int main(){
//FileOpen();
    scanf("%s",str + 1);
    int len = strlen(str + 1);

    nxt[0] = -1;
    int j = 0;
    R(i,2,len){
        while(j != -1 && str[j + 1] != str[i]) j = nxt[j];
        nxt[i] = ++j;
    }

    f[1] = 1;
    j = 0;
    R(i,2,len){
        f[i] = i;
        if(pos[f[nxt[i]]] >= i - nxt[i]) f[i] = f[nxt[i]];
        pos[f[i]] = i;
    }

    printf("%d", f[len]);
    return 0;
}