UVA1328 Period & SP263 PERIOD - Period

UVA1328 Period
对于每一个前缀$i$,$i-nexxt_i$即为最小循环节.证明在上一篇里.

这里要判断整除,否则就不行.
(代码可能有点不同,因为这题双倍经验,两道题输入不尽相同)
$Code:$

#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 ;
}

const int N = 1e6 + 100 ;

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

signed main (int argc , char * argv[]) {
    T = rint () ;
    while ( T -- ) {
        n = rint () ; if ( ! n ) break ;
        printf ("Test case #%lld\n" , ++ cnt ) ;
        MEM ( nxt , 0 ) ; 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 ;
        }
        rep ( i , 2 , n ) {
            int j = nxt[i] ; if ( i % ( i - j ) ) continue ;
            if ( j ) printf ("%lld %lld\n" , i , i / ( i - j ) ) ;
        }
        putchar(10);
    }
    system ("pause") ; return 0 ;
}

SP263 PERIOD - Period
双倍经验,除了输入一模一样.