fzu1901 kmp

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：找前缀后缀相同的地方
题解：kmp 的next数组（我居然没想到@。@）想成最小环了。。。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007

using namespace std;

const int N=1000000+5,maxn=1000000+5,inf=1e9+5;

int Next[N],slen,plen;
string str,ptr;

void getnext()
{
    Next[0]=-1;
    int k=-1;
    for(int i=1;i<slen;i++)
    {
        while(k>-1&&str[k+1]!=str[i])k=Next[k];
        if(str[k+1]==str[i])k++;
        Next[i]=k;
    }
}
bool kmp()
{
    int k=-1;
    for(int i=0;i<plen;i++)
    {
        while(k>-1&&str[k+1]!=ptr[i])k=Next[k];
        if(str[k+1]==ptr[i])k++;
        if(k==slen-1)return 1;
    }
    return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t,n,cnt=0;
    cin>>t;
    while(t--){
        cin>>str;
        slen=str.size();
        getnext();
        for(int i=0;i<slen;i++)
            cout<<Next[i]<<" ";
        cout<<endl;
        vector<int>q;
        int k=Next[slen-1],loop=slen-Next[slen-1]-1;
        while(k!=-1){
            q.push_back(slen-1-k);
            k=Next[k];
        }
        q.push_back(slen);
     /*   cout<<"Case #"<<++cnt<<": "<<q.size()<<endl;
        for(int i=0;i<q.size();i++)
        cout<<q[i]<<(i==q.size()-1 ? "\n":" ");*/
    }
    return 0;
}