E - Period poj1611（kmp 计算前缀循环节）

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结尾的前缀的循环节，关于kmp的next数组的性质，我前面的文章有介绍

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline 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<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> 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');
}
template<class T>inline 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<int,int> 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(i<len)
    {
        if(j==-1||s[i]==s[j])
            nex[++i]=++j;
        else
            j=nex[j];
    }
}
int main()
{
    //#ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    //freopen("zccccc.txt","w",stdout);
    //#endif // ONLINE_JUDGE
    int n,m,i,j,t,k;
    int T;
    int cas=1;
    while(read(n)&&n)
    {
        printf("Test case #%d\n",cas++);
        scanf("%s",a);
        int len=strlen(a);
        getnext(a);
        for(i=0;i<=len;i++)
        {
            t=i-nex[i];
            if(i%t==0&&i/t>1)
            {
                printf("%d %d\n",i,i/t);
            }
        }
        printf("\n");
    }

    return 0;
}