LuoGu P4391 [BOI2009]Radio Transmission 无线传输

P4391 [BOI2009]Radio Transmission 无线传输
这算是一点(next)数组的妙用吧...题目要求最小循环节(不要求恰好重复完成).先说结论,答案是(n-next_n),为什么呢?
根据(next)数组的定义,我们可以知道对于(i),(1)到(next_i)和(next_{i-next_i+1})到(i)是相同的,并且并不存在更大的(next)值满足这个条件.
所以:当(i-next_i)能整除(i)时,(1~i-next_i)就是最小循环节.因为这个题目不要求恰好重复完成,所以不需要判断整除,直接输出即可.
(Code:)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T 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 << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

template < class T >
    inline void write (T x) {
       static T stk[100] , top = 0 ;
       if ( x == 0 ) { putchar ('0') ; return ; }
       if ( x < 0 ) { x = - x ; putchar ( '-' ) ; }
       while ( x ) { stk[++top] = x % 10 ; x /= 10 ; }
       while ( top ) { putchar ( stk[top--] + '0' ) ; }
       putchar ('
') ;
    }

const int N = 1e6 + 100 ;

int n , nxt[N] ;
char s[N] ;

signed main (int argc , char * argv[]) {
    n = rint () ; scanf ("%s" , s + 1 ) ;
    int j = 0 ; rep ( i , 2 , n ) {
        while ( j && s[i] != s[j+1] ) j = nxt[j] ;
        if ( s[i] == s[j+1] ) ++ j ; nxt[i] = j ;
    }
    printf ("%lld
" , n - nxt[n] ) ;
    system ("pause") ; return 0 ;
}