FZU - 1901 Period II(kmp所有循环节)

Problem Description

For each prefix with length P of a given string S,if

S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

 Input

Input contains multiple cases.

The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

 Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.

 Sample Input

4

ooo

acmacmacmacmacma

fzufzufzuf

stostootssto

 Sample Output

Case #1: 3

1 2 3

Case #2: 6

3 6 9 12 15 16

Case #3: 4

3 6 9 10

Case #4: 2

9 12

 

题意:求出一个字符串的所有可能循环节的长度

思路:kmp的运用，题目的意思不是很清楚 j=next[j] 就是次大的匹配个数

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int nextt[1000007];
void getnext(string s){
    nextt[1]=0; int len=s.length(); 
    for(int i=2,j=0;i<=len;i++){
        while(j>0&&s[i-1]!=s[j]) j=nextt[j];
        if(s[i-1]==s[j]) j++;
        nextt[i]=j;
    }
}
int main(){
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    int w=0;
    while(t--){
        string s;
        cin>>s;
        getnext(s);
        int len=s.length();
        queue<int> q;
        int j=nextt[len];
        while(j>0){
            q.push(j);
            j=nextt[j];
        }
        q.push(0);
        bool f=1;
        cout<<"Case #"<<++w<<": "<<q.size()<<endl;
        while(!q.empty()){
            int temp=q.front();
            q.pop();
            if(f){
                cout<<len-temp;
                f=0;
            }else{
                cout<<" "<<len-temp;
            }
        }
        cout<<endl;
    } 
    return 0;
}