zoukankan      html  css  js  c++  java
  • LA3026

    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 AK , 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
    
    
    KMP模板题
    
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <string>
    #include <algorithm>
    #include <queue>
    #include <map>
    using namespace std;
    
    const int maxn = 1000000+10;
    int next[maxn];
    int n;
    string str;
    
    vector<int> find(string T,string P){
        int n = T.size(),m = P.size();
        vector<int> vn;
        getNext(P,next);
        for(int i = 0,j = 0; i < n; i++){
            while(j&&P[j] != T[i]) j = next[j];
            if(P[i]==T[j]) j++;
            if(j==m) vn.push_back(i-m+1);
        }
        return vn;
    
    }
    
    void getFail(string st,int *f){
        int m = st.size();
        next[0] = 0;
        next[1] = 0;
        int len = 1;
        for(int i = 1; i < m; i++){
            int j = next[i];
            while(j && st[i] != st[j]) j = next[j];
            if(st[i]==st[j]){
                next[i+1] = j+1;
            }else{
                next[i+1] = 0;
            }
        }
    }
    int main(){
        int T = 1;
        while(~scanf("%d",&n) && n){
            cin >> str;
            printf("Test case #%d
    ",T++);
            getFail(str,next);
            for(int i = 2; i <=  n; i++){
                if(next[i]> 0&&i%(i-next[i])==0){
                    cout<<i<<" "<<i/(i-next[i])<<endl;
                }
            }
            cout<<endl;
        }
        return 0;
    }
    


    版权声明:本文博客原创文章。博客,未经同意,不得转载。

  • 相关阅读:
    Oracle之内存结构(SGA、PGA)
    Android添加快捷方式(Shortcut)到手机桌面
    Android悬浮窗实现 使用WindowManager
    Android闹钟 AlarmManager的使用
    JavaScript学习13 JavaScript中的继承
    Android Content Provider Guides
    Android存储访问及目录
    ReflectUitls类的编写和对反射机制的解析
    Java File类总结和FileUtils类
    JavaScript学习12 JS中定义对象的几种方式
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/4732435.html
Copyright © 2011-2022 走看看