zoukankan      html  css  js  c++  java
  • kuangbin专题十六 KMP&&扩展KMP HDU1358 Period

    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.

    InputThe 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.
    OutputFor 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


    知道循环节怎么求,然后从头到尾处理Next[i]就可以了


     1 #include<stdio.h>
     2 #include<string.h>
     3 int Next[1000010],n;
     4 char p[1000010];
     5 
     6 void prekmp() {
     7     int i,j;
     8     j=Next[0]=-1;
     9     i=0;
    10     while(i<n) {
    11         while(j!=-1&&p[i]!=p[j]) j=Next[j];
    12         Next[++i]=++j;
    13     }
    14 }
    15 
    16 int main() {
    17     int Case=0;
    18     //freopen("out","r",stdin);
    19     while(~scanf("%d",&n)) {
    20         if(n==0) break;
    21         scanf("%s",p);
    22         prekmp();
    23         printf("Test case #%d
    ",++Case);
    24         for(int i=2;i<=n;i++) {
    25             int l=i-Next[i];
    26            // printf("%d***%d
    ",l,Next[i]);
    27             if(i%l==0&&i/l>1) printf("%d %d
    ",i,i/l);
    28         }
    29         printf("
    ");
    30     }
    31 }




  • 相关阅读:
    EF单实对应多表
    阻止保存要求重新创建表的更改
    Jenkins 修改主目录正解 workspace
    JNLP文件具体说明编辑
    Jenkins新建节点,启动方式没有“通过Java Web启动代理”选项怎么办?
    Entity Framework:“无法加载指定的元数据资源
    Java环境路径配置--转载
    趣谈StateServer在Web Garden,Web Farm下的使用
    【译文】漫谈ASP.NET中的Session
    好记心不如烂笔头之JQuery学习,第四章
  • 原文地址:https://www.cnblogs.com/ACMerszl/p/10266356.html
Copyright © 2011-2022 走看看