UVA1328 Period
对于每一个前缀(i),(i-next_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
" , ++ 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
" , 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.