E - Period poj1611（kmp 計算前綴循環節）

E - Period poj1611（kmp 計算前綴循環節）

E - Period
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status

Description
For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.


Input
The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) ? the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.


Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.


Sample Input

3
aaa
12
aabaabaabaab

0

 

t=i-next[i];表示已i結尾的前綴的循環節，關於kmp的next數組的性質，我前面的文章有介紹

 

#include
#include
#include
#include
#include
//#include
using namespace std;
templateinline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
templateinline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1000001;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair P;
#define bug printf("---\n");
#define mod  1000000007
int nex[maxn];
char a[maxn];

void getnext(char*s)
{
    int i=0,j=-1;
    nex[0]=-1;
    int len=strlen(s);
    while(i1)
            {
                printf("%d %d\n",i,i/t);
            }
        }
        printf("\n");
    }

    return 0;
}