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  • poj 2406 Power Strings KMP

    Power Strings

    Time Limit: 20 Sec  Memory Limit: 256 MB

    题目连接

    http://poj.org/problem?id=2406

    Description

    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

    题意

    找最小循环节的循环次数

    题解:

    KMP搞一搞就好了

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 4000001
    #define mod 10007
    #define eps 1e-9
    int Num;
    char CH[20];
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    /*
    
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    */
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    inline void P(int x)
    {
        Num=0;if(!x){putchar('0');puts("");return;}
        while(x>0)CH[++Num]=x%10,x/=10;
        while(Num)putchar(CH[Num--]+48);
        puts("");
    }
    //**************************************************************************************
    
    char s[maxn];
    int p[maxn];
    int main()
    {
        while(scanf("%s",s+1)!=EOF){
            if(s[1]=='.')
            break;
        int len=strlen(s+1);
        int j=0;
        for(int i=2;i<=len;i++)
        {
            while(j>0&&s[j+1]!=s[i])
                j=p[j];
            if(s[j+1]==s[i])
                j++;
            p[i]=j;
        }
        int tmp=0;
        int first=1;
        if(len%(len-p[len]))
            puts("1");
        else
            printf("%d
    ",len/(len-p[len]));
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4484646.html
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