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  • [kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher :G

    [kuangbin带你飞]专题十六 KMP & 扩展KMP & Manacher

    G - Power Strings

    POJ - 2406

    题目:

    Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
    Input
    Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
    Output
    For each s you should print the largest n such that s = a^n for some string a.
    Sample Input
    abcd
    aaaa
    ababab
    .
    
    Sample Output
    1
    4
    3
    
    Hint
    This problem has huge input, use scanf instead of cin to avoid time limit exceed.
    题意:求字符串的循环节个数
    思路:要注意如果该字符串如果不满足都是循环组成的则输出-1,若是都是循环节组成的,那么n-nextt[n]就是循环节长度,用总长度除以这个长度就是循环节长度
     
    // 
    // Created by HJYL on 2019/8/15.
    //
    #include <iostream>
    #include <vector>
    #include <map>
    #include <string>
    #include <queue>
    #include <stack>
    #include <set>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <cstdlib>
    using namespace std;
    const int maxn=1e7+10;
    char str[maxn];
    int nextt[maxn];
    void getnextt()
    {
        int i=0,j=-1;
        nextt[0]=-1;
        int n=strlen(str);
        while(i<n)
        {
            if(j==-1||str[i]==str[j])
            {
                i++,j++;
                if(str[i]!=str[j])
                    nextt[i]=j;
                else
                    nextt[i]=nextt[j];
            } else
                j=nextt[j];
        }
    }
    int main()
    {
        //freopen("C:\Users\asus567767\CLionProjects\untitled\text","r",stdin);
        while(~scanf("%s",str))
        {
            if(str[0]=='.')
                break;
            int len=strlen(str);
            getnextt();
            if(len%(len-nextt[len])==0)
             printf("%d
    ",len/(len-nextt[len]));
            else
                printf("1
    ");
        }
        return 0;
    }
     
     
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  • 原文地址:https://www.cnblogs.com/Vampire6/p/11360900.html
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