zoukankan      html  css  js  c++  java
  • hdu1358Period

    Problem 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 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
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    
    using namespace std;
    
    char s[1100000];
    int pn[1100000];
    int nextt[1100000];
    
    
    int main()
    {
        int l,z = 0,i,j,k;
        while(scanf("%d",&l)!=EOF,l)
        {
            printf("Test case #%d
    ",++z);
            scanf("%s",s+1);
            nextt[1] = 0;
            pn[1] = 1;
            for(i = 2;i<=l;i++)
            {
                int t = nextt[i-1];
                while(t&&s[i]!=s[t+1]) t = nextt[t];
                if(s[i] == s[t+1]) t++;
                nextt[i] = t;
                if(t == 0) pn[i] = 1;
                else if(i-t == t/pn[t]) pn[i] = pn[t]+1,printf("%d %d
    ",i,pn[i]);
                else pn[i] = 1;
            }
            puts("");
        }
    }

    注:hdu用next这个变量名会编译错误

  • 相关阅读:
    poj 1017 Packets
    hdu 1058 The sum problem
    HDU 1205 吃糖果
    Hdu 1004 Let the Balloon Rise map解决方法
    poj 1700 贪心算法(1)
    大数计算器
    大整数的加减乘除取模
    【单调队列】poj 2823 Sliding Window
    【单调队列】bzoj 1407 [HAOI2007]理想的正方形
    【单调队列+二分查找】bzoj 1012: [JSOI2008]最大数maxnumber
  • 原文地址:https://www.cnblogs.com/wos1239/p/4397992.html
Copyright © 2011-2022 走看看