zoukankan      html  css  js  c++  java
  • poj1961 Period

    Description

    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 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

    题意是给定s,枚举每个s的前缀,判断是不是由一个串重复多次组成。如果是输出前缀长度和这个串最多自我复制几次
    这题还是跟bzoj1355一样……枚举每个i和i-next[i],如果能整除就输出。
    #include<cstdio>
    #include<cstring>
    int next[1000010];
    char s[1000010];
    int cnt,l,len,j;
    inline void pre()
    {
    	memset(next,0,sizeof(next));
    	j=0;
    	for (int i=2;i<=l;i++)
    	{
    		while (j>0 && s[j+1]!=s[i])j=next[j];
    		if (s[j+1]==s[i])j++;
    		next[i]=j;
    	}
    }
    int main()
    {
    	while (scanf("%d",&l)&&l)
    	{
    		scanf("%s",s+1);
    		pre();
    		printf("Test case #%d
    ",++cnt);
    		for(int i=1;i<=l;i++)
    		{
    			len=i-next[i];
    			if (i%len==0&&next[i])
    			  printf("%d %d
    ",i,i/len);
    		}
    		printf("
    ");
    	}
    	return 0;
    }
    

      

    ——by zhber,转载请注明来源
  • 相关阅读:
    Spring面试,IoC和AOP的理解
    WEB打印(jsp版)
    Spring事务管理机制的实现原理-动态代理
    spring面试题
    oracle PLSQL基础学习
    oracle创建表空间
    WM_CONCAT字符超过4000的处理办法
    Oracle 数据泵使用详解
    Oracle 数据泵详解
    linux下启动oracle
  • 原文地址:https://www.cnblogs.com/zhber/p/4162979.html
Copyright © 2011-2022 走看看