Problem C FZU 1901

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


自己写了一百分钟
没用KMP
主要是错了没发现
改了以后结果对了
只是超时

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    int T;
    cin>>T;
    int t = 0;
    while(T--)
    {
        t++;
        string s;
        cin>>s;
        int len =1;
        int num =0;
        int* length = new int[s.length()];
        while(len<s.length())
        {
            string sub = s.substr(0,len);
            int index = 0;
            bool ok = true;
            while(index<s.length())
            {

                string temp = s.substr(index,len);
                if(temp.length()<len)
                {
                    sub = sub.substr(0,temp.length());
                }
                if(sub.compare(temp)!=0)
                {
                    ok = false;

                    break;
                }else{
                    index+=len;
                }
            }


            if(ok==true)
            {
                length[num] = len;
                num++;
            }
            len++;
        }
        num++;
        cout<<"Case #"<<t<<": "<<num<<endl;
        for(int i =0;i<num-1;i++)
        {
            cout<<length[i]<<" ";
        }
        cout<<s.length()<<endl;
    }
    return 0;
}

怎么才能不超时呢

我们考虑next(len)，令t=next(len); next(len)有什么含义？ str[1…t]=str[len-t+1…len] 那么，长度为len-next(len)的前缀显然是符合题意的。 接下来我们应该去考虑谁？ t=next( next(len) ); t=next( next (next(len) ) ); 一直下去直到t=0,每个符合题意的前缀长是len-t。

没想到

我想的是从小到大

忘记了递归

#include <iostream>
#include<string>
using namespace std;

#define N 1000100
int a[N],next[N];
int m;
string s;
void getNext(){
    next[0]=0;
    next[1]=0;
    for(int i=1;i<m;i++){
        int j=next[i];
        while(s.at(j)!=s.at(i)&&j) j=next[j];
        if(s.at(j)==s.at(i)) next[i+1]=j+1;
        else next[i+1]=0;
    }
}

int main()
{
    int T,count;
    cin.sync_with_stdio(false);
    cin>>T;
    for(int i=1;i<=T;i++){
        cin>>s;
        m=s.length(),count=0;
        int t=m;
        getNext();

        while(next[m])
        {
            a[count++]=t-next[m];
            m=next[m];
        }

        cout<<"Case #"<<i<<": "<<count+1<<endl;
        for(int i=0;i<count;i++) cout<<a[i]<<' ';
        cout<<t<<endl;
    }
    return 0;
}

其实不如跳过写下一题