HDU 1358 && HDU 3746 && POJ 2406 最小循环节 (KMP)

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4486    Accepted Submission(s): 2164

Problem 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

 

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

 

http://www.cnblogs.com/wuyiqi/archive/2012/01/06/2314078.html

最小循环节还是要自己多去模拟仔细体会

 

HDU 1358 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
const int MAXN=1000000+5;
char str[MAXN];
int nex[MAXN];
int n;
int Case=0;

void getnext()
{
    int j,k;
    j=0,k=-1,nex[0]=-1;
    while(j<n)
    {
        if(k==-1 || str[j]==str[k])
            nex[++j]=++k;
        else
            k=nex[k];
    }
}
int solve()
{
    getnext();
    int res;
    printf("Test case #%d
",++Case);
    for(int i=1;i<=n;i++)
    {
        res=i-nex[i];
        if(i%res==0 && i/res>1)
            printf("%d %d
",i,i/res);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    Case=0;
    while(scanf("%d",&n) && n)
    {
        scanf("%s",str);
        solve();
        printf("
");
    }
    return 0;
}

HDU 3746

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
const int MAXN=100000+5;
char str[MAXN];
int nex[MAXN];
int n;

void getnext()
{
    int j=0,k=-1;
    nex[0]=-1;
    while(j<n)
    {
        if(k==-1 || str[j]==str[k])
            nex[++j]=++k;
        else
            k=nex[k];
    }
}
void solve()
{
    getnext();

    int res=n-nex[n];
    if(n%res==0 && n/res>1)
        printf("0
");
    else
        printf("%d
",res-nex[n]%res);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int kase;
    scanf("%d",&kase);
    while(kase--)
    {
        scanf("%s",str);
        n=strlen(str);
        solve();
    }
    return 0;
}

POJ 2406

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
const int MAXN=1000000+5;
char str[MAXN];
int nex[MAXN];
int len;
void getnext()
{
    int j,k;
    j=0,k=-1,nex[0]=-1;
    while(j<len)
    {
        if(k==-1 || str[j]==str[k])
            nex[++j]=++k;
        else
            k=nex[k];
    }
}
int main()
{
    while(scanf("%s",str)!=EOF)
    {
        if(str[0]=='.') break;

        len=strlen(str);
        getnext();

        int res=len-nex[len];

        if(len%res==0 && len/res>1) printf("%d
",len/res);
        else printf("1
");
    }
    return 0;
}